Question

find a function f whose graph is a parabola that has the given vertex and that passes through the indicated point.
vertex (4, −5); point(5, 2)
f(x) =

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if a ball is thrown directly upward with a velocity of 28 ft/s, its height (in feet) after t seconds is given by y = 28t − 16t (nearest whole number.)
ft

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Given vertex \((4, -5)\), so \( h = 4 \), \( k = -5 \). Then \( f(x) = a(x - 4)^2 - 5 \).

Step2: Substitute the point \((5, 2)\)

Substitute \( x = 5 \), \( f(x) = 2 \) into the equation: \( 2 = a(5 - 4)^2 - 5 \). Simplify: \( 2 = a(1)^2 - 5 \) → \( 2 = a - 5 \).

Step3: Solve for \( a \)

Add 5 to both sides: \( a = 2 + 5 = 7 \).

Step4: Write the function

Substitute \( a = 7 \) back into the vertex form: \( f(x) = 7(x - 4)^2 - 5 \). Expand if needed: \( f(x) = 7(x^2 - 8x + 16) - 5 = 7x^2 - 56x + 112 - 5 = 7x^2 - 56x + 107 \). But the vertex form is also acceptable.

Answer:

\( 7(x - 4)^2 - 5 \) (or \( 7x^2 - 56x + 107 \))