Question

write an equation for the parabola that has the given vertex and passes through the given point.
vertex
(6,0)
point
(8, 4)
$f(x) = ?(x + \\ )^2 + \\ $

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 \((6,0)\), so \(h = 6\), \(k=0\). Then the equation becomes \( f(x)=a(x - 6)^{2}+0=a(x - 6)^{2} \). But the given form is \( f(x)=a(x + b)^{2}+c \), note that \(x - 6=x+(- 6)\), so \(b=-6\), \(c = 0\) for now, and we need to find \(a\).

Step2: Substitute the point into the equation

The parabola passes through \((8,4)\). Substitute \(x = 8\), \(f(x)=4\) into \( f(x)=a(x - 6)^{2} \):
\(4=a(8 - 6)^{2}\)
\(4=a(2)^{2}\)
\(4 = 4a\)

Step3: Solve for \(a\)

Divide both sides by 4: \(a=\frac{4}{4}=1\)

Answer:

\(f(x)=1(x + (-6))^{2}+0\) (or in the boxed form as per the question's blanks: the first blank is \(1\), the second is \(-6\), the third is \(0\))