Question

question 20
consider the function $h(x) = -2x^2 + 20x - 51$ graph $h(x)$
what is the vertex of $h$?
what is the equation of the line of symmetry of $h$?
$h$ has a select an answer of
the $x$-intercept(s) of $h$ is/are
the $y$-intercept of $h$ is
for this question give all rational answers as fractions or integers and all irrational answers rounded to 2 decimal places.
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Explanation:

Step1: Find vertex x-coordinate

For quadratic $ax^2+bx+c$, $x=-\frac{b}{2a}$.
$a=-2, b=20$, so $x=-\frac{20}{2(-2)} = 5$

Step2: Find vertex y-coordinate

Substitute $x=5$ into $h(x)$.
$h(5)=-2(5)^2+20(5)-51 = -50+100-51 = -1$

Step3: Identify line of symmetry

Line of symmetry is $x=$ vertex x-coordinate.
$x=5$

Step4: Determine max/min value

Since $a=-2<0$, parabola opens downward, so it has a maximum at vertex y-value.
Max value = $-1$

Step5: Find x-intercepts

Set $h(x)=0$, solve $-2x^2+20x-51=0$. Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

$$\begin{align*} x&=\frac{-20\pm\sqrt{20^2-4(-2)(-51)}}{2(-2)}\\ &=\frac{-20\pm\sqrt{400-408}}{-4}\\ &=\frac{-20\pm\sqrt{-8}}{-4}\\ &=\frac{-20\pm2i\sqrt{2}}{-4}\\ &=5\pm\frac{i\sqrt{2}}{2} \end{align*}$$

These are complex, so no real x-intercepts.

Step6: Find y-intercept

Set $x=0$, solve $h(0)$.
$h(0)=-2(0)^2+20(0)-51 = -51$

Answer:

• Vertex of $h$: $(5, -1)$
• Equation of the line of symmetry: $x=5$
• $h$ has a maximum of $-1$
• The $x$-intercept(s) of $h$: None (no real intercepts)
• The $y$-intercept of $h$: $-51$

(For graphing: Plot the vertex $(5,-1)$, the y-intercept $(0,-51)$, and use the line of symmetry $x=5$ to plot the mirror point of the y-intercept $(10,-51)$, then draw a downward-opening parabola through these points.)