Question

which statements are true about the graph of the function $h(x) = -5x^2 + 60x - 200$? select three options.
the axis of symmetry is the line $x = -6$.
the graph of $h(x)$ will not intersect the graph of the parent function, $f(x) = x^2$.
the vertex of the graph is at $(6, -20)$.
the parabola has a maximum.
the value of $k$, when the equation is written in vertex form, is $-200$.

Explanation:

Step1: Find axis of symmetry

For $ax^2+bx+c$, axis is $x=-\frac{b}{2a}$.
$a=-5, b=60$, so $x=-\frac{60}{2(-5)} = 6$

Step2: Check vertex coordinates

Substitute $x=6$ into $h(x)$:
$h(6)=-5(6)^2+60(6)-200 = -5(36)+360-200 = -180+360-200 = -20$
Vertex is $(6,-20)$

Step3: Check parabola direction

Since $a=-5<0$, parabola opens downward, so it has a maximum.

Step4: Check intersection with $f(x)=x^2$

Set $x^2=-5x^2+60x-200$:
$6x^2-60x+200=0$, discriminant $\Delta=(-60)^2-4(6)(200)=3600-4800=-1200<0$, so no real intersections.

Step5: Check vertex form $k$ value

Vertex form is $h(x)=a(x-h)^2+k$, vertex $(h,k)=(6,-20)$, so $k=-20$, not $-200$.

Answer:

• The graph of $h(x)$ will not intersect the graph of the parent function, $f(x) = x^2$.
• The vertex of the graph is at $(6, -20)$.
• The parabola has a maximum.