Question

8-58. graph ( y = x^2 - 8x + 7 ) and label its vertex, ( x )-intercepts, and ( y )-intercepts.

Explanation:

Step1: Find the vertex (x-coordinate)

For quadratic $y=ax^2+bx+c$, $x=-\frac{b}{2a}$.
Here $a=1$, $b=-8$, so:
$$x=-\frac{-8}{2(1)}=4$$

Step2: Find vertex (y-coordinate)

Substitute $x=4$ into the equation:
$$y=(4)^2-8(4)+7=16-32+7=-9$$
Vertex: $(4, -9)$

Step3: Find y-intercept (x=0)

Substitute $x=0$ into the equation:
$$y=0^2-8(0)+7=7$$
Y-intercept: $(0, 7)$

Step4: Find x-intercepts (y=0)

Solve $x^2-8x+7=0$. Factor:
$$(x-1)(x-7)=0$$
Set each factor to 0:
$x-1=0 \implies x=1$; $x-7=0 \implies x=7$
X-intercepts: $(1, 0)$ and $(7, 0)$

Step5: Plot and connect points

Plot vertex $(4,-9)$, y-intercept $(0,7)$, x-intercepts $(1,0)$ and $(7,0)$, then draw a symmetric parabola opening upward.

Answer:

• Vertex: $\boldsymbol{(4, -9)}$
• X-intercepts: $\boldsymbol{(1, 0)}$ and $\boldsymbol{(7, 0)}$
• Y-intercept: $\boldsymbol{(0, 7)}$

(Graph: A upward-opening parabola passing through the above points, symmetric about the vertical line $x=4$)