# SOLUTION: Suppose you wanted to design a parabolic arch (a parabola opening downward) as a trellis in your garden. This would mean that the first term of the quadratic equation, the term wit

Algebra ->  Algebra  -> Graphs -> SOLUTION: Suppose you wanted to design a parabolic arch (a parabola opening downward) as a trellis in your garden. This would mean that the first term of the quadratic equation, the term wit      Log On

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 Click here to see ALL problems on Graphs Question 89510: Suppose you wanted to design a parabolic arch (a parabola opening downward) as a trellis in your garden. This would mean that the first term of the quadratic equation, the term with the variable raised to the second power, would have to be negative. An engineering friend suggests that the following quadratic function might be aesthetically pleasing. h(x) = -5x2/64 + 4 This function gives the height of the arch (in feet) at any horizontal distance x (in feet) from the centerline of the arch. The trellis (parabola) could be supported by vertical posts at intervals across the inside of the arch. These posts could be placed at 1, 2, 3, 4, 5, 6, and 7 feet on both sides of the centerline (the axis of the parabola). Therefore, for any distance (x-value) the height of a post supporting the arch at that point could be determined from the above function. 1. When drawing the graph of this parabola, where is its axis located in the coordinate plane? 2. Where would the arch touch the ground? (i.e. How far from the axis of the parabola?) Suppose you wanted to design a parabolic arch (a parabola opening downward) as a trellis in your garden. This would mean that the first term of the quadratic equation, the term with the variable raised to the second power, would have to be negative. An engineering friend suggests that the following quadratic function might be aesthetically pleasing. h(x) = -5x^2/64 + 4 This function gives the height of the arch (in feet) at any horizontal distance x (in feet) from the centerline of the arch. The trellis (parabola) could be supported by vertical posts at intervals across the inside of the arch. These posts could be placed at 1, 2, 3, 4, 5, 6, and 7 feet on both sides of the centerline (the axis of the parabola). Therefore, for any distance (x-value) the height of a post supporting the arch at that point could be determined from the above function. 1. When drawing the graph of this parabola, where is its axis located in the coordinate plane? Please show the graph 2. Where would the arch touch the ground? (i.e. How far from the axis of the parabola?) 3. What is the maximum height of the arch? 4. What are the lengths of the posts at the distances of 1, 2, 3, . . . feet from the axis? a. 1 foot from axis b. 2 feet from axis c. 3 feet from axis d. 4 feet from axis e. 5 feet from axis f. 6 feet from axis g. 7 feet from axis a. = b. = c. = d. = e. = f. = g. = I have submitted this problem several times and now I think I figured out what I was doing wrong please help me. Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!1. the axis of symmetry for a graph of the form ax^2+bx+c is x=-b/2a ... since b=0 in this case, the axis of symmetry is x=0 (the y-axis) 2. the arch touches the ground when h(x)=0 ... 0=-5x^2/64+4 ... -4=-5x^2/64 ... -256=-5x^2 ... 51.2=x^2 ... x=+-7.16 (approx.) 3. the maximum height occurs at the axis of symmetry, where x=0 ... h=-5(0)^2/64+4 ... h=4 4. a thru g ... to find the height of the posts, plug the distance (x) values into the function and evaluate