Question

write an equation for the parabola that has the given vertex and passes through the given point.
vertex (4, -3) point (7, 0)
f(x) = (1)/(?) (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 \((4,-3)\), so \(h = 4\), \(k=-3\). Substitute into the formula: \( f(x)=a(x - 4)^2-3 \).

Step2: Substitute the point \((7,0)\) to find \(a\)

Substitute \(x = 7\) and \(f(x)=0\) into \( f(x)=a(x - 4)^2-3 \):
\[
0=a(7 - 4)^2-3
\]
\[
0=a(3)^2-3
\]
\[
0 = 9a-3
\]
Add 3 to both sides: \( 9a=3 \), then \(a=\frac{3}{9}=\frac{1}{3}\).

Step3: Rewrite the equation in the given form

The vertex form is \( f(x)=\frac{1}{3}(x - 4)^2-3 \). Now, rewrite \((x - 4)\) as \((x+(-4))\). So the equation is \( f(x)=\frac{1}{3}(x + (-4))^2+(-3) \).

Answer:

For the first box (the coefficient of the square term's denominator) is \(3\), the second box (inside the \(x+\square\)) is \(-4\), and the third box (the constant term) is \(-3\). So the equation is \( f(x)=\frac{1}{3}(x + (-4))^2+(-3) \), so the filled values are \(3\), \(-4\), \(-3\) (in the order of the boxes: the denominator of the fraction is \(3\), the term inside \(x+\square\) is \(-4\), and the last constant is \(-3\)).