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Zorluk: ZorQuadratic Functions and Graphs

In the xyxy-plane, the graph of the quadratic function f(x)=−x2+bx+cf(x) = -x^2 + bx + c, where bb and cc are constants, has its vertex at (h,k)(h, k). The function gg is defined by g(x)=f(x−3)+4g(x) = f(x - 3) + 4. The graph of gg passes through the origin (0,0)(0,0), and its vertex lies on the line y=xy = x in the first quadrant. What is the value of f(0)f(0)?

  1. A
    -1
  2. -7Cevap
  3. C
    -3
  4. D
    -21

Cevap

-7
The correct answer is −7-7. Representing f(x)f(x) in vertex form as f(x)=−(x−h)2+kf(x) = -(x-h)^2 + k tells us that translating the function 3 units right and 4 units up shifts the vertex from (h,k)(h, k) to (h+3,k+4)(h+3, k+4). Since this vertex lies on the line y=xy=x, we have h+3=k+4h+3 = k+4, or k=h−1k = h-1. Since the vertex of gg lies in the first quadrant, its coordinates must be positive, meaning h>−3h > -3 and k>−4k > -4. Using the fact that the graph of gg passes through the origin, we have g(0)=0  ⟹  f(−3)+4=0  ⟹  f(−3)=−4g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4. Substituting x=−3x = -3 into the vertex form of f(x)f(x) gives −(−3−h)2+k=−4  ⟹  (h+3)2−k=4-(-3-h)^2 + k = -4 \implies (h+3)^2 - k = 4. Substituting k=h−1k = h-1 yields (h+3)2−(h−1)=4  ⟹  h2+5h+6=0(h+3)^2 - (h-1) = 4 \implies h^2 + 5h + 6 = 0, which factors into (h+2)(h+3)=0(h+2)(h+3) = 0. The solution h=−3h = -3 is discarded because it places the vertex of gg at (0,0)(0,0), which is not in the first quadrant. Therefore, h=−2h = -2 and k=−3k = -3. Calculating f(0)f(0) gives −h2+k=−(−2)2+(−3)=−7-h^2 + k = -(-2)^2 + (-3) = -7.

Adım Adım Çözüm

1
Determine the vertex form of f(x)f(x) and the vertex of g(x)g(x).
Since f(x)=−x2+bx+cf(x) = -x^2 + bx + c has its vertex at (h,k)(h, k), its vertex form is f(x)=−(x−h)2+kf(x) = -(x - h)^2 + k. The function g(x)=f(x−3)+4g(x) = f(x - 3) + 4 represents a horizontal shift of ff by 3 units to the right and a vertical shift by 4 units up. Thus, the vertex of gg is (h+3,k+4)(h + 3, k + 4).
Understanding translations allows us to write the coordinates of the new vertex in terms of the original vertex variables.
2
Use the line y=xy = x and quadrant constraints to find a relation between hh and kk.
Since the vertex of gg lies on the line y=xy = x, we set its coordinates equal: h+3=k+4h + 3 = k + 4, which simplifies to k=h−1k = h - 1. Additionally, because the vertex is in the first quadrant, we must have h+3>0h + 3 > 0 and k+4>0k + 4 > 0.
The geometric placement of the vertex on the line y=xy = x constrains its coordinate values.
3
Apply the condition that the graph of gg passes through the origin.
g(0)=0  ⟹  f(−3)+4=0  ⟹  f(−3)=−4g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4. Substituting x=−3x = -3 into the vertex form of f(x)f(x) gives −(−3−h)2+k=−4-(-3 - h)^2 + k = -4, which simplifies to (h+3)2−k=4(h + 3)^2 - k = 4.
The point (0,0)(0,0) lying on the graph of gg provides an equation to solve for the vertex parameters.
4
Substitute k=h−1k = h - 1 into the quadratic equation and solve for hh.
(h+3)2−(h−1)=4  ⟹  h2+6h+9−h+1=4  ⟹  h2+5h+6=0(h + 3)^2 - (h - 1) = 4 \implies h^2 + 6h + 9 - h + 1 = 4 \implies h^2 + 5h + 6 = 0. Factoring gives (h+2)(h+3)=0(h + 2)(h + 3) = 0, so h=−2h = -2 or h=−3h = -3.
Substituting the linear relation into the quadratic equation isolates the variable hh.
5
Filter the solutions using the first quadrant constraint and calculate f(0)f(0).
If h=−3h = -3, then k=−4k = -4, giving the vertex of gg at (0,0)(0, 0), which is not in the first quadrant. If h=−2h = -2, then k=−3k = -3, giving the vertex of gg at (1,1)(1, 1), which is in the first quadrant. Thus, h=−2h = -2 and k=−3k = -3. The value of f(0)f(0) is f(0)=−(0−h)2+k=−h2+k=−(−2)2+(−3)=−4−3=−7f(0) = -(0 - h)^2 + k = -h^2 + k = -(-2)^2 + (-3) = -4 - 3 = -7.
The quadrant condition uniquely determines the correct vertex coordinates, allowing us to find the y-intercept of the original function.

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Quadratic Functions and Graphs
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