Question

write the standard form of the quadratic function that has the indi result.
vertex: (2, 1); point: (4, -7)
f(x) = -2x² + 8x - 7 ×
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Explanation:

Step1: Recall vertex form of quadratic

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, \( h = 2 \), \( k = 1 \), so \( f(x) = a(x - 2)^2 + 1 \).

Step2: Substitute the point \((4, -7)\)

Substitute \( x = 4 \) and \( f(x) = -7 \) into the vertex form: \(-7 = a(4 - 2)^2 + 1\). Simplify: \(-7 = a(2)^2 + 1\) → \(-7 = 4a + 1\).

Step3: Solve for \( a \)

Subtract 1 from both sides: \(-8 = 4a\). Divide by 4: \( a = -2 \).

Step4: Write vertex form and expand

Now, \( f(x) = -2(x - 2)^2 + 1 \). Expand \((x - 2)^2 = x^2 - 4x + 4\), so \( f(x) = -2(x^2 - 4x + 4) + 1 = -2x^2 + 8x - 8 + 1 = -2x^2 + 8x - 7 \). Wait, but let's check the vertex of \( f(x) = -2x^2 + 8x - 7 \). The x-coordinate of vertex is \( -\frac{b}{2a} = -\frac{8}{2(-2)} = 2 \), and \( f(2) = -2(4) + 16 - 7 = -8 + 16 - 7 = 1 \), which matches the vertex. And \( f(4) = -2(16) + 32 - 7 = -32 + 32 - 7 = -7 \), which matches the point. Wait, the original answer was marked wrong, but according to calculations, it's correct? Wait, maybe a typo in the system. Wait, let's recheck. Wait, the standard form is \( ax^2 + bx + c \), and we derived \( f(x) = -2x^2 + 8x - 7 \), which has vertex (2,1) and passes through (4,-7). So maybe the marking was incorrect. But according to the steps, the answer should be \( -2x^2 + 8x - 7 \).

Answer:

\( f(x) = -2x^2 + 8x - 7 \)