Question

1. suppose you knew $f(x)=x^{2}-8x - 9=(x + 1)(x - 9)$. the zeros are ____. find the vertex using the zeros.
2. similar to one of the problems in wiley consider the following function: $y=x(x - 4)-6(x - 4)$

a. write the function in standard form, $y = ax^{2}+bx + c$, and identify $a,b,c$ and the vertical intercept.
b. write the function in factored form, $y=a(x - r)(x - s)$, and identify $r,s$ and the zeros.
c. write the function in vertex form, $y=a(x - h)^{2}+k$, and identify the vertex.
d. use all the information above to sketch a graph of the function. make sure to very clearly show all of the features listed above on your graph. hint: you might want to scale the vertical axis by units of 2 in order to fit the value of $c$ on the graph.

Explanation:

Step1: Find zeros and vertex for \(f(x)=x^{2}-8x - 9\)

For zeros, set \((x + 1)(x - 9)=0\), so \(x=-1,9\). For vertex \(x\) - coordinate, use \(h=\frac{-1 + 9}{2}=4\), then find \(y\) - coordinate.

Step2: Expand to standard form for \(y=x(x - 4)-6(x - 4)\)

Multiply out terms: \(y=x^{2}-4x-6x + 24=x^{2}-10x + 24\), identify \(a,b,c\) and vertical - intercept.

Step3: Factor the function

Factor out common factor \((x - 4)\) to get \(y=(x - 4)(x - 6)\), identify \(r,s\) and zeros.

Step4: Complete the square for vertex form

Rewrite \(x^{2}-10x + 24\) as \((x - 5)^{2}-1\) to find vertex.

Step5: Sketch the graph

Use zeros, vertical intercept and vertex to draw the parabola.

Answer:

1.

Zeros:

The zeros of \(f(x)=x^{2}-8x - 9=(x + 1)(x - 9)\) are \(x=-1\) and \(x = 9\).
The \(x\) - coordinate of the vertex of a parabola \(y = a(x - r)(x - s)\) is \(h=\frac{r + s}{2}\). Here \(r=-1\) and \(s = 9\), so \(h=\frac{-1+9}{2}=4\).
Substitute \(x = 4\) into \(y=x^{2}-8x - 9\): \(y=4^{2}-8\times4-9=16-32 - 9=-25\). The vertex is \((4,-25)\).

2.

a.

First, expand \(y=x(x - 4)-6(x - 4)\):
\[

$$\begin{align*} y&=x^{2}-4x-6x + 24\\ y&=x^{2}-10x + 24 \end{align*}$$

\]
Here \(a = 1\), \(b=-10\), \(c = 24\). The vertical - intercept is the value of \(y\) when \(x = 0\), so when \(x = 0\), \(y=24\).

b.

Factor \(y=x(x - 4)-6(x - 4)\) as \(y=(x - 4)(x - 6)\). Here \(a = 1\), \(r = 4\), \(s = 6\), and the zeros are \(x = 4\) and \(x = 6\).

c.

Complete the square for \(y=x^{2}-10x + 24\).
\[

$$\begin{align*} y&=x^{2}-10x+24\\ y&=(x^{2}-10x+25)-25 + 24\\ y&=(x - 5)^{2}-1 \end{align*}$$

\]
The vertex is \((5,-1)\).

d.

• Zeros: Mark the points \((4,0)\) and \((6,0)\) on the \(x\) - axis.
• Vertical intercept: Mark the point \((0,24)\) on the \(y\) - axis.
• Vertex: Mark the point \((5,-1)\).

The parabola opens upward (since \(a = 1>0\)). Sketch a smooth curve passing through these points.