Question

an arch at a new amusement park is constructed with two vertical walls and a parabolic arch top, where x represents the horizontal distance in feet and f(x) represents the height in feet. the y - axis represents the vertical left edge of the arch. since the arch and its base are symmetrical, complete the equation for the line containing the vertical, right edge of the arch. x = \square. (graph: a parabola with points (-10, 0), (0, 20), (30, 45), (70, 0). the x - axis ranges from -20 to 90, y - axis from -30 to 60.)

Explanation:

Step1: Find the axis of symmetry

The parabola has roots at \( x = -10 \) and \( x = 70 \). The axis of symmetry of a parabola is the midpoint between its roots. The formula for the midpoint between two points \( x_1 \) and \( x_2 \) is \( x=\frac{x_1 + x_2}{2} \). Substituting \( x_1=-10 \) and \( x_2 = 70 \), we get \( x=\frac{-10 + 70}{2}=\frac{60}{2}=30 \). So the axis of symmetry is \( x = 30 \).

Step2: Determine the right vertical edge

The left vertical edge is the \( y \)-axis (\( x = 0 \)). Since the arch is symmetric about \( x = 30 \), the distance from the left edge (\( x = 0 \)) to the axis of symmetry (\( x = 30 \)) is \( 30 \) units. So the right vertical edge should be \( 30 \) units to the right of the axis of symmetry? Wait, no. Wait, the left edge is \( x = 0 \), and the axis of symmetry is \( x = 30 \). Wait, actually, the distance from \( x=-10 \) to \( x = 70 \) is \( 80 \) units, so the midpoint is at \( x = 30 \) (as we calculated). The left vertical wall: wait, the problem says "the \( y \)-axis represents the vertical left edge of the arch". Wait, but the root is at \( x=-10 \) and \( x = 70 \). Wait, maybe I misread. Wait, the left vertical edge is \( x = 0 \) (the \( y \)-axis), and we need to find the right vertical edge. Since the parabola is symmetric about its axis ( \( x = 30 \) ), the distance from the left edge ( \( x = 0 \) ) to the axis ( \( x = 30 \) ) is \( 30 \) units. So the right edge should be \( 30 \) units to the right of the axis? Wait, no. Wait, let's think again. The axis of symmetry is \( x = 30 \). The left vertical edge is \( x = 0 \). The distance between \( x = 0 \) and \( x = 30 \) is \( 30 \). So the right vertical edge should be \( 30 \) units to the right of \( x = 30 \)? Wait, no, that can't be. Wait, maybe the left edge is \( x = 0 \), and the parabola is symmetric about \( x = 30 \), so the right edge is \( x = 60 \)? Wait, no, let's check the roots. The roots are at \( x=-10 \) and \( x = 70 \). The midpoint is \( x=\frac{-10 + 70}{2}=30 \), correct. The left vertical edge is \( x = 0 \). The distance from \( x = 0 \) to \( x = 30 \) is \( 30 \). So the right vertical edge should be \( x = 30+30 = 60 \)? Wait, but the root is at \( x = 70 \). Wait, maybe I made a mistake. Wait, the problem says "two vertical walls and a parabolic arch top". The vertical walls are vertical lines, so they are parallel to the \( y \)-axis, so their equations are \( x = a \) and \( x = b \). The left wall is \( x = 0 \) (the \( y \)-axis). The parabola is symmetric about its axis ( \( x = 30 \) ), so the distance between \( x = 0 \) and \( x = 30 \) is \( 30 \), so the distance between \( x = 30 \) and \( x = b \) should also be \( 30 \), so \( b = 30 + 30 = 60 \). Wait, but the root is at \( x = 70 \). Wait, maybe the vertical walls are not at the roots, but at the base. Wait, the problem says "the arch and its base are symmetrical". The base is between the two vertical walls? Wait, maybe the left vertical wall is \( x = 0 \), and the right vertical wall is \( x = 60 \), because the axis of symmetry is \( x = 30 \), so the distance from \( x = 0 \) to \( x = 30 \) is \( 30 \), so \( x = 30+30 = 60 \). Let's verify. If the left wall is \( x = 0 \) and the right wall is \( x = 60 \), then the base is from \( x = 0 \) to \( x = 60 \), and the parabola is symmetric about \( x = 30 \). Let's check the point \( (70, 0) \): wait, \( x = 70 \) is a root, but maybe the vertical walls are not at the roots. Wait, the problem says "two vertical walls and a parabolic arch top", so the arch is between the two vertical walls. The left wall is \( x = 0 \) ( \( y \)-axis), and we need to find the right wall. Since the arch is symmetric, the axis of symmetry is the midpoint between the two vertical walls. So if the left wall is \( x = 0 \), and the axis of symmetry is \( x = 30 \), then the right wall is \( x = 60 \) (because \( 0 + 60 = 60 \), midpoint is \( 30 \) ). That makes sense. So the equation of the right vertical edge is \( x = 60 \). Wait, but let's check with the graph. The point \( (70, 0) \) is a root, but maybe that's the end of the parabola, not the wall. Wait, the problem says "two vertical walls and a parabolic arch top", so the arch is between the two walls, and the parabola goes from the left wall, up, and down to the right wall? Wait, maybe I was wrong earlier. Let's re-express. The axis of symmetry is \( x = 30 \). The left vertical wall is \( x = 0 \). The distance from \( x = 0 \) to \( x = 30 \) is \( 30 \), so the right vertical wall is \( x = 30 + 30 = 60 \). So the equation of the right vertical wall is \( x = 60 \). Wait, but let's check the graph. The point \( (70, 0) \) is on the parabola, but maybe the vertical walls are at \( x = 0 \) and \( x = 60 \), and the parabola extends beyond the walls? No, the problem says "two vertical walls and a parabolic arch top", so the arch is between the walls. So the left wall is \( x = 0 \), right wall is \( x = 60 \), and the parabola is between them, symmetric about \( x = 30 \). So the equation of the right vertical edge (wall) is \( x = 60 \).

Answer:

\( 60 \)