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In the xy-plane the graph of the function q is a parabola. The graph intersects the x-axis at (−1, 0) and (r, 0). If the vertex of q occurs at the point (2, 4), what is the value of r ?

A) 0

B) 3

C) 4

D) 5

**Comments**

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Since q is a function and its graph is a parabola, it follows that q is a quadratic function and the parabola is symmetric about the vertical line through its vertex.

Thus, the x-coordinate of the vertex (2, 4) is the average of the x-coordinate of the two x-intercepts (–1, 0) and (r, 0).

That is,

2 = (-1 + r)/2

It follows that 4 = –1 + r,

r = 5

**Answer: D**

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The vertex form of a parabola is given by the equation:

y = a(x - h)^{2} + k

where point (h, k) is the vertex of the parabola, and a is a constant which indicates the direction of the parabola.

Given vertex = (2, 4),

y = a(x - 2)^{2} + 4

Next, find the value of a.

At point (-1, 0), x = -1 and y = 0

0 = a(-1 - 2)^{2} + 4

0 = a(-3)^{2 }+ 4

0 = 9a + 4

9a = -4

a = -4/9

Substitute the value of a into the equation

y = -4/9 (x - 2)^{2} + 4

At point (r, 0), x = r and y = 0

0 = -4/9 (r - 2)^{2} + 4

4/9 (r - 2)^{2} = 4

multiply both sides by 9

4(r - 2)^{2} = 36

divide both sides by 4

(r - 2)^{2} = 9

Take the square root of both sides

r - 2 = 3

r = 3 + 2

r = 5

**Answer: D**

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