Question

monday 10/6 and tuesday 10/7
for problems 5 - 7: convert the fu

1. $f(x)=x^{2}-10x + 24$

vertex form: ____________
vertex: ____________

Explanation:

Step1: Complete the square

For the quadratic function \(f(x)=x^{2}-10x + 24\), the coefficient of \(x\) is \(- 10\). Half of it is \(\frac{-10}{2}=-5\), and its square is \(25\).
\[

$$\begin{align*} f(x)&=x^{2}-10x + 24\\ &=(x^{2}-10x+25)-25 + 24 \end{align*}$$

\]

Step2: Simplify to vertex - form

Using the perfect - square formula \((a - b)^2=a^{2}-2ab + b^{2}\) (here \(a = x\) and \(b = 5\)), we get \(f(x)=(x - 5)^{2}-1\). The vertex form of a quadratic function is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex of the parabola.

Step3: Identify the vertex

For the function \(y=(x - 5)^{2}-1\), comparing with \(y=a(x - h)^{2}+k\), we have \(h = 5\) and \(k=-1\). So the vertex is \((5,-1)\).

Answer:

Vertex Form: \(f(x)=(x - 5)^{2}-1\)
Vertex: \((5,-1)\)