answer：Alright, so I've got this problem here where I need to find the points of intersection between a quadratic function and a linear function. The quadratic is given by ( f(x) = x^2 - 6x + 8 ) and the linear function is ( g(x) = 2x - 4 ). The first part asks for the points where these two functions intersect, which I think means solving for x when ( f(x) = g(x) ). Then, the second part wants to know for which values of x the quadratic function is greater than the linear one, so I need to figure out where ( f(x) > g(x) ).Okay, starting with the first part. To find the points of intersection, I need to set the two functions equal to each other:( x^2 - 6x + 8 = 2x - 4 )Hmm, so I can rearrange this equation to bring all terms to one side. Let me subtract ( 2x ) and add 4 to both sides to get:( x^2 - 6x + 8 - 2x + 4 = 0 )Wait, that doesn't seem right. Let me check. If I subtract ( 2x ) and add 4, it should be:Left side: ( x^2 - 6x + 8 - 2x + 4 )Right side: 0But combining like terms on the left:( x^2 - 6x - 2x + 8 + 4 = x^2 - 8x + 12 = 0 )Okay, so the quadratic equation to solve is ( x^2 - 8x + 12 = 0 ). I can try factoring this. Looking for two numbers that multiply to 12 and add up to -8. Let's see, factors of 12 are 1 & 12, 2 & 6, 3 & 4. Since the middle term is -8x, both numbers should be negative because their product is positive and their sum is negative. So, -6 and -2 multiply to 12 and add to -8. Perfect.So, factoring:( (x - 6)(x - 2) = 0 )Setting each factor equal to zero:( x - 6 = 0 ) => ( x = 6 )( x - 2 = 0 ) => ( x = 2 )So, the points of intersection are at x = 2 and x = 6. To find the corresponding y-values, I can plug these back into either function. I'll choose the linear function ( g(x) = 2x - 4 ) because it's simpler.For x = 2:( g(2) = 2*2 - 4 = 4 - 4 = 0 )For x = 6:( g(6) = 2*6 - 4 = 12 - 4 = 8 )So, the points are (2, 0) and (6, 8). That seems straightforward.Now, moving on to the second part: determining for which values of x the quadratic function ( f(x) ) is greater than the linear function ( g(x) ). So, we need to solve the inequality:( x^2 - 6x + 8 > 2x - 4 )Again, let's bring all terms to one side:( x^2 - 6x + 8 - 2x + 4 > 0 )Simplify:( x^2 - 8x + 12 > 0 )Wait, this is the same quadratic as before! So, we have ( x^2 - 8x + 12 > 0 ). We already factored this as ( (x - 6)(x - 2) > 0 ).To solve this inequality, I can analyze the sign of the quadratic expression. The quadratic opens upwards because the coefficient of ( x^2 ) is positive. The roots are at x = 2 and x = 6, so the parabola crosses the x-axis at these points. The expression will be positive outside the interval [2, 6] and negative inside.So, the inequality ( (x - 6)(x - 2) > 0 ) holds true when x < 2 or x > 6.Therefore, ( f(x) > g(x) ) when x is less than 2 or greater than 6.Let me just double-check this. If I pick a value less than 2, say x = 0:( f(0) = 0 - 0 + 8 = 8 )( g(0) = 0 - 4 = -4 )8 > -4, which is true.If I pick a value between 2 and 6, say x = 4:( f(4) = 16 - 24 + 8 = 0 )( g(4) = 8 - 4 = 4 )0 < 4, so f(x) is not greater than g(x) here.If I pick a value greater than 6, say x = 7:( f(7) = 49 - 42 + 8 = 15 )( g(7) = 14 - 4 = 10 )15 > 10, which is true.So, my conclusion seems correct.In summary, the points of intersection are at (2, 0) and (6, 8), and the quadratic function is greater than the linear function when x is less than 2 or greater than 6.**Final Answer**1. The points of intersection are boxed{(2, 0)} and boxed{(6, 8)}.2. The quadratic function is greater than the linear function for ( x < 2 ) and ( x > 6 ), which can be written as boxed{(-infty, 2) cup (6, infty)}.

answer：First, I need to establish the relationship between the number of Avant-garde pieces (A) and Vintage pieces (V). According to the problem, the number of Avant-garde pieces should be 5 more than twice the number of Vintage pieces. This can be expressed as A = 2V + 5.Next, I know that the total number of pieces in the collection should be 30. This means that the sum of Avant-garde and Vintage pieces is 30, which can be written as A + V = 30.To find the values of A and V, I'll substitute the expression for A from the first equation into the second equation. This gives me (2V + 5) + V = 30. Combining like terms, I get 3V + 5 = 30. Subtracting 5 from both sides, I have 3V = 25. Dividing both sides by 3, I find that V = 25/3.Finally, I'll substitute V back into the first equation to find A. A = 2*(25/3) + 5 = 50/3 + 15/3 = 65/3. Therefore, there are 65/3 Avant-garde pieces and 25/3 Vintage pieces in the collection.

answer：Alright, so I have these two problems to solve related to setting up a nightclub with a nostalgic 80s theme. Let me tackle them one by one.Starting with the first problem about the lighting system. The goal is to figure out how many spotlights and disco balls to buy without exceeding the 5000 budget, while also having twice as many spotlights as disco balls. Hmm, okay, let's break this down.First, let's define some variables. Let me call the number of disco balls "d" and the number of spotlights "s". According to the problem, I want twice as many spotlights as disco balls. So that translates to s = 2d. Got that.Now, the cost for each spotlight is 150, and each disco ball is 250. The total budget is 5000. So the total cost equation would be 150s + 250d = 5000. Since I know that s is twice d, I can substitute s with 2d in the equation. Let me write that out:150*(2d) + 250d = 5000Simplifying that, 150*2 is 300, so:300d + 250d = 5000Adding those together, 300d + 250d is 550d. So:550d = 5000To find d, I need to divide both sides by 550.d = 5000 / 550Let me compute that. 5000 divided by 550. Hmm, 550 goes into 5000 how many times? Let's see, 550*9 is 4950, which is just 50 less than 5000. So, 5000 / 550 = 9 with a remainder of 50. So, that's 9 and 50/550, which simplifies to 9 and 1/11. But since I can't buy a fraction of a disco ball, I need to round down to 9 disco balls.Wait, but if I round down, let me check the total cost. If d=9, then s=2*9=18.Total cost would be 150*18 + 250*9.Calculating 150*18: 150*10=1500, 150*8=1200, so total 2700.250*9=2250.Adding those together: 2700 + 2250 = 4950. That's within the budget, leaving 50 unused.But wait, could I maybe buy 10 disco balls? Let's check. If d=10, then s=20.Total cost would be 150*20 + 250*10 = 3000 + 2500 = 5500, which is over the budget by 500. So that's not possible.So, 9 disco balls and 18 spotlights is the maximum I can buy without exceeding the budget. That seems to be the solution.Moving on to the second problem about the dance floor. The dance floor is 400 square feet, so it's a square with sides of 20 feet each because 20*20=400. I need to create a large square in the center using neon tiles, leaving a 2-foot border around it. Each neon tile covers 1 square foot, so the number of neon tiles needed will be the area of the large square.First, let me visualize this. The dance floor is a 20x20 square. There's a 2-foot border around the edges, so the large square in the center will be smaller. To find the dimensions of the large square, I need to subtract twice the border width from each side.So, the side length of the large square would be 20 - 2*2 = 20 - 4 = 16 feet. Therefore, the area of the large square is 16*16=256 square feet. Since each tile is 1 square foot, that means I need 256 neon tiles.Wait, let me double-check that. If the border is 2 feet wide, then on each side of the dance floor, both left and right, top and bottom, there's a 2-foot border. So, the length of the inner square is 20 - 2 - 2 = 16 feet. Yep, that's correct. So, the area is 16^2=256. So, 256 neon tiles are needed.Alternatively, I can think of it as the total area minus the border area. The total area is 400. The border is a frame around the inner square. The border's area can be calculated as the total area minus the inner area. But since I already know the inner area is 256, the border area would be 400 - 256 = 144. But since the question only asks for the number of neon tiles, which is the inner area, 256 is the answer.Just to make sure, let me consider the width of the border. If the border is 2 feet, then the inner square is reduced by 2 feet on each side. So, starting from 20, subtract 2 from both sides, so 20 - 4 = 16. Yep, that's consistent.So, to recap, for the first problem, purchasing 9 disco balls and 18 spotlights uses 4950 of the budget, which is the maximum possible without exceeding 5000. For the second problem, the number of neon tiles needed is 256.**Final Answer**1. You should purchase boxed{18} spotlights and boxed{9} disco balls.2. The number of neon tiles needed is boxed{256}.

answer：First, I need to determine the number of panel discussions on the first day. The problem states that there are twice as many panel discussions as film screenings, and there are 5 film screenings planned. So, I'll calculate the number of panel discussions by multiplying the number of film screenings by 2.Next, I'll move on to the second and third days. The organizer wants the total number of film screenings to be 4 more than the number of panel discussions over these two days. I'll represent the number of events on the second day as (2x^2 - 3x + 5) and on the third day as (x^2 + x + 4). To find the total number of events over the three days, I'll add the events from each day together.Since the total number of events must not exceed 50, I'll set up an inequality where the sum of the events from all three days is less than or equal to 50. This will give me a quadratic inequality to solve for (x). I'll simplify the inequality by combining like terms and then solve for (x) using the quadratic formula or by factoring if possible.Finally, I'll determine the appropriate value of (x) that satisfies the inequality, ensuring that the total number of events remains within the desired limit.

answer：Okay, so I have this problem about a watercolor artist who is trying to replicate a historical artist's technique. The problem has two parts, and I need to solve both. Let me take them one at a time.Starting with the first part: The height ( h(x) ) of each triangular layer is given by the polynomial function ( h(x) = 2x^3 - 3x^2 + x - 5 ). The artist wants the height of the 3rd layer to be 10 units. Wait, hold on, is that the 3rd layer or the layer when ( x = 3 )? Hmm, the wording says "the height of the 3rd layer," so I think that means when ( x = 3 ). But the question is asking for the value of ( x ) when the height is 10 units. So, maybe I misread. Let me check again.It says, "the artist wants the height of the 3rd layer to be 10 units." So, does that mean that when ( x = 3 ), ( h(x) = 10 )? Or is it that the 3rd layer (which would be ( x = 3 )) should have a height of 10? Or is it that the height of the 3rd layer is 10, and we need to find ( x )? Hmm, the wording is a bit confusing. Let me parse it again."The artist wants the height of the 3rd layer to be 10 units. Find the value of ( x ) when the height of the triangular layer is 10 units."Wait, so maybe the 3rd layer is when ( x = 3 ), but the height there is given by ( h(3) ). But the artist wants that height to be 10. So, perhaps they need to adjust something? Or maybe it's just a translation error. Alternatively, maybe the 3rd layer is when ( x = 3 ), but the height is 10, so we need to solve for ( x ) when ( h(x) = 10 ). Hmm.Wait, maybe the problem is saying that the height of the 3rd layer is 10, so ( h(3) = 10 ). But if ( h(x) = 2x^3 - 3x^2 + x - 5 ), then ( h(3) = 2*(27) - 3*(9) + 3 - 5 = 54 - 27 + 3 - 5 = 25. So, h(3) is 25, but the artist wants it to be 10. So, perhaps the polynomial is incorrect? Or maybe I'm misunderstanding.Wait, maybe the problem is just asking, given ( h(x) = 2x^3 - 3x^2 + x - 5 ), find the value of ( x ) such that ( h(x) = 10 ). So, regardless of the 3rd layer, just solve ( 2x^3 - 3x^2 + x - 5 = 10 ). That would make more sense. So, perhaps the mention of the 3rd layer is just context, but the actual question is to solve for ( x ) when ( h(x) = 10 ).Let me read the problem again:1. The height ( h(x) ) of each triangular layer is given by the polynomial function ( h(x) = 2x^3 - 3x^2 + x - 5 ), where ( x ) represents the layer number. The artist wants the height of the 3rd layer to be 10 units. Find the value of ( x ) when the height of the triangular layer is 10 units.So, it's saying that the artist wants the 3rd layer (x=3) to have a height of 10. But when x=3, h(3)=25, which is not 10. So, perhaps the artist needs to adjust the polynomial? Or maybe the problem is just to find x such that h(x)=10, regardless of the layer number.Wait, perhaps the problem is just to solve h(x)=10, regardless of the 3rd layer. Maybe the mention of the 3rd layer is just an example or context. Let me check the exact wording:"The artist wants the height of the 3rd layer to be 10 units. Find the value of ( x ) when the height of the triangular layer is 10 units."So, they want the 3rd layer's height to be 10, but the function is given as h(x)=2x^3 -3x^2 +x -5. So, perhaps they need to adjust the function? Or maybe it's a typo, and they meant the 3rd layer is when h(x)=10, so solve for x.Wait, maybe the problem is just to solve h(x)=10, regardless of x=3. Because if x=3 gives h=25, which is not 10, then the artist wants to find x such that h(x)=10.Alternatively, maybe the problem is saying that the 3rd layer (x=3) should have a height of 10, so they need to adjust the polynomial. But the polynomial is given, so perhaps the question is just to solve h(x)=10, regardless of x=3.I think the problem is just to solve h(x)=10, so let's proceed with that.So, equation: 2x^3 - 3x^2 + x - 5 = 10Subtract 10 from both sides: 2x^3 - 3x^2 + x - 15 = 0So, we have a cubic equation: 2x^3 - 3x^2 + x - 15 = 0We need to find the real roots of this equation.Let me try to factor this. Maybe rational root theorem can help. Possible rational roots are factors of 15 over factors of 2, so ±1, ±3, ±5, ±15, ±1/2, ±3/2, ±5/2, ±15/2.Let me test x=3: 2*(27) - 3*(9) + 3 -15 = 54 -27 +3 -15=15, not zero.x=5: 2*125 -3*25 +5 -15=250-75+5-15=165, not zero.x=1: 2 -3 +1 -15= -15, not zero.x= -1: -2 -3 -1 -15= -21, no.x=1/2: 2*(1/8) -3*(1/4) +1/2 -15= 1/4 - 3/4 +1/2 -15= (1 -3 +2)/4 -15=0/4 -15= -15, not zero.x=3/2: 2*(27/8) -3*(9/4) +3/2 -15= 27/4 - 27/4 + 3/2 -15= 0 + 3/2 -15= -13.5, not zero.x=5/2: 2*(125/8) -3*(25/4) +5/2 -15= 125/4 - 75/4 +5/2 -15= (125-75)/4 +5/2 -15= 50/4 +5/2 -15=12.5 +2.5 -15=0. So, x=5/2 is a root.Great, so (x - 5/2) is a factor. Let's perform polynomial division or use synthetic division.Let me use synthetic division with x=5/2.Coefficients: 2 | -3 | 1 | -15Bring down 2.Multiply 2 by 5/2: 5.Add to next coefficient: -3 +5=2.Multiply 2 by 5/2:5.Add to next coefficient:1 +5=6.Multiply 6 by 5/2=15.Add to last coefficient: -15 +15=0.So, the cubic factors as (x - 5/2)(2x^2 + 2x +6).Now, set each factor equal to zero.x -5/2=0 => x=5/2=2.52x^2 +2x +6=0Divide by 2: x^2 +x +3=0Discriminant: 1 -12= -11, so no real roots.Thus, the only real solution is x=5/2.So, x=2.5.But x represents the layer number, which should be a positive integer, right? Because layers are numbered 1,2,3,...But 2.5 is not an integer. Hmm, that's a problem.Wait, maybe I made a mistake in interpreting the problem. Let me check again.The problem says: "the artist wants the height of the 3rd layer to be 10 units. Find the value of ( x ) when the height of the triangular layer is 10 units."So, perhaps the artist is not necessarily using integer layer numbers? Or maybe the layers can be fractional? That seems odd, but perhaps in the context of the problem, it's acceptable.Alternatively, maybe the problem is just to solve h(x)=10, regardless of x being an integer. So, the answer is x=2.5.But let me think again. The function is given as h(x)=2x^3 -3x^2 +x -5, where x is the layer number. So, x should be a positive integer, starting from 1,2,3,...But when x=3, h(3)=25, which is higher than 10. So, maybe the artist is using a different starting point? Or perhaps the function is supposed to model the height for any x, not necessarily integer.Alternatively, maybe the problem is just to solve h(x)=10, regardless of x being an integer. So, the answer is x=5/2.But let me check if x=2.5 is acceptable. The problem doesn't specify that x must be an integer, just that it's the layer number. So, perhaps in this context, x can be any real number, representing the layer in a continuous sense.Therefore, the solution is x=5/2.So, part 1 answer is x=5/2.Now, moving on to part 2.The area ( A ) of each triangular layer is given by the quadratic equation ( A = frac{1}{2}b^2 + 3b + 2 ). The artist wants to determine for which base length(s) the area of the triangle is 20 square units. Find the possible base lengths for which the area is 20 square units.So, we have A=20, so set up the equation:20 = (1/2)b² + 3b + 2Multiply both sides by 2 to eliminate the fraction:40 = b² + 6b +4Bring all terms to one side:b² +6b +4 -40=0 => b² +6b -36=0So, quadratic equation: b² +6b -36=0We can solve this using quadratic formula.b = [-6 ± sqrt(36 + 144)] /2Because discriminant D=6² -4*1*(-36)=36 +144=180So, sqrt(180)=sqrt(36*5)=6*sqrt(5)Thus, b=(-6 ±6√5)/2= (-6/2) ± (6√5)/2= -3 ±3√5So, the solutions are b= -3 +3√5 and b= -3 -3√5But since base length cannot be negative, we discard the negative solution.So, b= -3 +3√5Calculate that:√5≈2.236, so 3√5≈6.708Thus, b≈-3 +6.708≈3.708 unitsSo, the base length is approximately 3.708 units, but we can write it exactly as 3(√5 -1)Alternatively, factor 3: b=3(√5 -1)So, exact value is 3(√5 -1), approximate value is about 3.708.Therefore, the possible base length is 3(√5 -1) units.Let me double-check the calculations.Starting from A=20= (1/2)b² +3b +2Multiply by 2: 40= b² +6b +4Subtract 40: b² +6b -36=0Quadratic formula: b=(-6 ±sqrt(36 +144))/2=(-6 ±sqrt(180))/2=(-6 ±6√5)/2= -3 ±3√5Yes, that's correct. So, only positive solution is -3 +3√5, which is approximately 3.708.So, part 2 answer is b= -3 +3√5, or 3(√5 -1).I think that's it.

answer：Alright, so I have this problem about a drone's flight path modeled by a quadratic equation. The equation is given as ( y = -2x^2 + 8x + 3 ). There are two parts to the problem: first, finding the maximum height the drone reaches, and second, determining the range of horizontal distances ( x ) where the drone is at least 5 meters above the ground. Let me try to work through each part step by step.Starting with the first part: finding the maximum height. I remember that quadratic equations graph as parabolas, and since the coefficient of ( x^2 ) is negative (-2), the parabola opens downward. That means the vertex of the parabola will be its highest point, which is the maximum height in this context.To find the vertex of a quadratic equation in standard form ( y = ax^2 + bx + c ), the x-coordinate of the vertex can be found using the formula ( x = -frac{b}{2a} ). Once I have the x-coordinate, I can plug it back into the original equation to find the corresponding y-coordinate, which will be the maximum height.So, in this equation, ( a = -2 ), ( b = 8 ), and ( c = 3 ). Plugging into the vertex formula:( x = -frac{8}{2*(-2)} = -frac{8}{-4} = 2 ).Okay, so the x-coordinate of the vertex is 2 meters. Now, substituting ( x = 2 ) back into the equation to find the maximum height:( y = -2(2)^2 + 8(2) + 3 ).Calculating each term:- ( (2)^2 = 4 )- ( -2 * 4 = -8 )- ( 8 * 2 = 16 )- So, ( y = -8 + 16 + 3 )Adding those together:( -8 + 16 = 8 )( 8 + 3 = 11 )So, the maximum height is 11 meters. That seems straightforward. Let me just double-check my calculations to make sure I didn't make a mistake.Wait, let me recalculate ( y ) when ( x = 2 ):( y = -2*(2)^2 + 8*2 + 3 )( y = -2*4 + 16 + 3 )( y = -8 + 16 + 3 )( y = 8 + 3 = 11 ). Yep, that's correct.Alright, so part one is done. The maximum height is 11 meters.Moving on to part two: finding the range of ( x ) where the drone is at least 5 meters above the ground. So, we need to find all ( x ) such that ( y geq 5 ).Given the equation ( y = -2x^2 + 8x + 3 ), we can set up the inequality:( -2x^2 + 8x + 3 geq 5 ).To solve this inequality, I'll first subtract 5 from both sides to bring all terms to one side:( -2x^2 + 8x + 3 - 5 geq 0 )( -2x^2 + 8x - 2 geq 0 ).Now, we have a quadratic inequality: ( -2x^2 + 8x - 2 geq 0 ).I can simplify this by dividing all terms by -2. However, I remember that when multiplying or dividing an inequality by a negative number, the direction of the inequality sign flips. So, dividing both sides by -2:( x^2 - 4x + 1 leq 0 ).Now, the inequality is ( x^2 - 4x + 1 leq 0 ). To find the solution, I need to find the roots of the quadratic equation ( x^2 - 4x + 1 = 0 ) and then determine the intervals where the quadratic expression is less than or equal to zero.Using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -4 ), and ( c = 1 ):Discriminant ( D = (-4)^2 - 4*1*1 = 16 - 4 = 12 ).So, the roots are:( x = frac{4 pm sqrt{12}}{2} ).Simplify ( sqrt{12} ) as ( 2sqrt{3} ), so:( x = frac{4 pm 2sqrt{3}}{2} = 2 pm sqrt{3} ).Therefore, the roots are ( x = 2 + sqrt{3} ) and ( x = 2 - sqrt{3} ).Since ( sqrt{3} ) is approximately 1.732, the roots are approximately:- ( 2 + 1.732 = 3.732 )- ( 2 - 1.732 = 0.268 )So, the quadratic ( x^2 - 4x + 1 ) crosses the x-axis at approximately 0.268 and 3.732. Since the coefficient of ( x^2 ) is positive (1), the parabola opens upward. That means the quadratic expression ( x^2 - 4x + 1 ) is below or equal to zero between its roots.Therefore, the solution to ( x^2 - 4x + 1 leq 0 ) is ( 2 - sqrt{3} leq x leq 2 + sqrt{3} ).But wait, let me make sure I didn't make a mistake in the direction of the inequality. So, we started with ( y geq 5 ), which led us to ( -2x^2 + 8x - 2 geq 0 ), then dividing by -2 (which flips the inequality) to get ( x^2 - 4x + 1 leq 0 ). So, yes, the solution is between the roots.Therefore, the drone is at least 5 meters above the ground when ( x ) is between ( 2 - sqrt{3} ) and ( 2 + sqrt{3} ).But let me verify this by plugging in a value within the interval and outside the interval to see if the inequality holds.First, let's take ( x = 2 ), which is inside the interval. Plugging into ( y = -2x^2 + 8x + 3 ):( y = -2*(4) + 16 + 3 = -8 + 16 + 3 = 11 ), which is greater than 5. So, that's correct.Now, let's take ( x = 0 ), which is less than ( 2 - sqrt{3} approx 0.268 ). Plugging into the equation:( y = -2*(0)^2 + 8*(0) + 3 = 0 + 0 + 3 = 3 ), which is less than 5. So, that's correct.Similarly, take ( x = 4 ), which is greater than ( 2 + sqrt{3} approx 3.732 ). Plugging in:( y = -2*(16) + 32 + 3 = -32 + 32 + 3 = 3 ), which is also less than 5. So, that's correct.Therefore, the range where the drone is at least 5 meters above the ground is from ( 2 - sqrt{3} ) meters to ( 2 + sqrt{3} ) meters.But just to be thorough, let me also check the endpoints. At ( x = 2 - sqrt{3} ):( y = -2*(2 - sqrt{3})^2 + 8*(2 - sqrt{3}) + 3 ).First, compute ( (2 - sqrt{3})^2 = 4 - 4sqrt{3} + 3 = 7 - 4sqrt{3} ).So, ( y = -2*(7 - 4sqrt{3}) + 16 - 8sqrt{3} + 3 ).Calculating each term:- ( -2*7 = -14 )- ( -2*(-4sqrt{3}) = +8sqrt{3} )- ( 16 - 8sqrt{3} )- ( +3 )Combine all terms:( -14 + 8sqrt{3} + 16 - 8sqrt{3} + 3 ).Simplify:- ( -14 + 16 = 2 )- ( 8sqrt{3} - 8sqrt{3} = 0 )- ( 2 + 3 = 5 )So, ( y = 5 ) at ( x = 2 - sqrt{3} ). Similarly, at ( x = 2 + sqrt{3} ), the same calculation would yield ( y = 5 ). So, the endpoints are indeed where the drone is exactly 5 meters above the ground, and between these points, it's higher than 5 meters.Therefore, summarizing:1. The maximum height is 11 meters.2. The drone is at least 5 meters above the ground for horizontal distances between ( 2 - sqrt{3} ) meters and ( 2 + sqrt{3} ) meters.I think that covers both parts of the problem. I tried to be methodical and check each step to make sure I didn't make any calculation errors. It's easy to mix up signs or miscalculate when dealing with quadratics, so double-checking is always a good idea.**Final Answer**1. The maximum height the drone reaches is boxed{11} meters.2. The drone is at least 5 meters above the ground for horizontal distances between boxed{2 - sqrt{3}} meters and boxed{2 + sqrt{3}} meters.