Try the fastest way to create flashcards
Question

# 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

Verified
Step 1
1 of 2

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}

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

## Recommended textbook solutions

#### Algebra 2, California Edition

1st EditionISBN: 9780078659805Beatrice Moore Harris, Carter, Casey, Cuevas, Day, Hayek, Holliday, Marks
8,110 solutions

#### enVision Algebra 2

1st EditionISBN: 9780328931590 (1 more)Al Cuoco
3,573 solutions

#### Big Ideas Math Algebra 2: A Common Core Curriculum

1st EditionISBN: 9781608408405 (2 more)Boswell, Larson
5,067 solutions

#### Big Ideas Math: Algebra 2 A Common Core Curriculum

1st EditionISBN: 9781642088052Laurie Boswell, Ron Larson
5,067 solutions