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Zorluk: OrtaQuadratic Functions and Graphs

The graph of the quadratic function ff in the xyxy-plane has xx-intercepts at (2,0)(-2, 0) and (8,0)(8, 0). If the maximum value of f(x)f(x) is 2525, what is the value of f(0)f(0)?

Cevap: 16

Cevap

16
The axis of symmetry of the quadratic function ff lies halfway between the xx-intercepts x=2x = -2 and x=8x = 8, which is at x=2+82=3x = \frac{-2 + 8}{2} = 3. Since the function has a maximum value of 2525, this maximum must occur at the vertex, giving the vertex coordinates (3,25)(3, 25). In vertex form, the function is f(x)=a(x3)2+25f(x) = a(x - 3)^2 + 25. Substituting the xx-intercept (8,0)(8, 0) into the function yields 0=a(83)2+250 = a(8 - 3)^2 + 25, which simplifies to 25a=2525a = -25, or a=1a = -1. Therefore, the equation of the function is f(x)=(x3)2+25f(x) = -(x - 3)^2 + 25. Evaluating this at x=0x = 0 gives f(0)=(03)2+25=9+25=16f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16.

Adım Adım Çözüm

1
Find the xx-coordinate of the vertex (axis of symmetry)
x=3x = 3
The axis of symmetry of a parabola is located exactly halfway between its xx-intercepts: x=2+82=3x = \frac{-2 + 8}{2} = 3.
2
Determine the vertex coordinates
(3,25)(3, 25)
The maximum value of the quadratic function occurs at its vertex, so the yy-coordinate of the vertex is the maximum value 2525.
3
Write the vertex form of the quadratic function
f(x)=a(x3)2+25f(x) = a(x - 3)^2 + 25
The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
4
Solve for the leading coefficient aa
a=1a = -1
Substitute the xx-intercept (8,0)(8, 0) into the vertex form: 0=a(83)2+25    25a=25    a=10 = a(8 - 3)^2 + 25 \implies 25a = -25 \implies a = -1.
5
Find the value of f(0)f(0)
f(0)=16f(0) = 16
Substitute x=0x = 0 into the function: f(0)=(03)2+25=9+25=16f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16.

Anahtar Kavram

Using xx-intercepts and the maximum value to determine the vertex and equation of a quadratic function.
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