QUESTION IMAGE
Question
what is the equation of the quadratic function with a vertex at (2,-25) and an x-intercept at (7,0)?
○ $f(x) = (x - 2)(x - 7)$
○ $f(x) = (x + 2)(x + 7)$
○ $f(x) = (x - 3)(x + 7)$
○ $f(x) = (x + 3)(x - 7)$
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, \( h = 2 \), \( k = -25 \), so \( f(x) = a(x - 2)^2 - 25 \).
Step2: Use x - intercept to find a
We know the x - intercept is \((7, 0)\), so substitute \( x = 7 \) and \( f(x) = 0 \) into the vertex form:
\( 0 = a(7 - 2)^2 - 25 \)
\( 0 = 25a - 25 \)
\( 25a = 25 \)
\( a = 1 \)
Step3: Expand the vertex form
Now that \( a = 1 \), the function is \( f(x)=(x - 2)^2 - 25 \). Expand \((x - 2)^2 - 25\):
\( (x - 2)^2 - 25=x^{2}-4x + 4-25=x^{2}-4x - 21 \)
Factor \( x^{2}-4x - 21 \): We need two numbers that multiply to - 21 and add to - 4. The numbers are - 7 and 3. So \( x^{2}-4x - 21=(x + 3)(x - 7) \)
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D. \( f(x)=(x + 3)(x - 7) \) (assuming the last option is D, if we follow the order of options: first option A, second B, third C, fourth D)