Question

match each quadratic function to its graph.
$f(x) = -x^2 + 7x - 6 = -(x - 1)(x - 6)$
$g(x) = -x^2$
two blue boxes with $f(x) = -x^2 + 7x - 6$ and $g(x) = -x^2$ respectively, followed by two graphs on grids: left graph has a parabola opening down with vertex around (4,6), crossing x-axis at ~1 and ~6, y-intercept negative; right graph has a parabola opening down with vertex at (0,0), symmetric about y-axis

Explanation:

Step1: Analyze $g(x) = -x^2$

This quadratic function has vertex at $(0,0)$ (since $g(0)=0$), and it opens downward (negative leading coefficient). It only touches the x-axis at $x=0$.

Step2: Analyze $f(x) = -x^2 +7x -6$

First, find x-intercepts: set $f(x)=0$, so $-(x-1)(x-6)=0$, giving $x=1$ and $x=6$. The y-intercept is $f(0)=-6$. The vertex x-coordinate is $\frac{-b}{2a}=\frac{-7}{2(-1)}=3.5$, and $f(3.5)=-(3.5)^2+7(3.5)-6=6.25$, so vertex at $(3.5, 6.25)$.

Step3: Match to graphs

The right graph has vertex at $(0,0)$ and no other x-intercepts, matching $g(x)$. The left graph has x-intercepts at 1 and 6, y-intercept at -6, matching $f(x)$.

Answer:

• $f(x) = -x^2 + 7x - 6$ matches the left graph (with x-intercepts at 1 and 6, vertex above the x-axis, y-intercept at -6)
• $g(x) = -x^2$ matches the right graph (with vertex at (0,0), opening downward, touching x-axis only at origin)