Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the given equation. Then find the length of the latus rectum and graph the parabola.
$y=3 x^2-24 x+50$

Solution

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Answered 2 years ago

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Rewrite in standard form.

\begin{aligned} y&=3x^2-24x+50\\ y&=3\left[x^2-8x+\left(\frac{b}{2}\right)^2\right]+50-3\left(\frac{b}{2}\right)^2\\ y&=3\left[x^2-8x+\left(\frac{-8}{2}\right)^2\right]+50-3\left(\frac{-8}{2}\right)^2\\ y&=3\left[x^2-8x+\left(16\right)\right]+50-3\left(16\right)\\ y&=3(x-4)^2+2 \end{aligned}

Thus the standard form is

\begin{aligned} y&=3(x-4)^2+2 \end{aligned}

Therefore, $\text{vertex}=(4,2)$ $\text{axis of symmetry}=4$

\begin{aligned}\text{focus}&=\left (h,k+\dfrac{1}{4a}\right)\\ &=\left (4,2+\dfrac{1}{4(3)}\right)\\ &=\left (4,2+\dfrac{1}{12}\right)\\ &=\left (4,\dfrac{24+1}{12}\right)\\ &=\left (4,\dfrac{25}{12}\right)\\ \end{aligned}

$\text{focus}=\left (4,2.08\right)\\$

\begin{aligned}\text{directrix}:y&=k-\dfrac{1}{4a}\\ &=2-\dfrac{1}{4(3)}\\ &=2-\frac{1}{12}\\ &=\frac{24-1}{12}\\ &=\frac{23}{12} \end{aligned}

$\text{directrix}=1.92$ $a>0= \text {opens upward}$

\begin{aligned}\text{lenght of latus rectum}&=\dfrac{1}{a}\text{ unit}\\&=\dfrac{1}{3} \end{aligned}

$\text{latus rectum}=\frac{1}{3} \text{ unit}$

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