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Expert-verified Found in: Page 423 ### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903 # Write an equation of the parabola that will be open to the left.

The vertex is $\left(3,-7\right)$. Focus is $\left(3,-\frac{111}{16}\right)$. Axis of symmetry is $x=3$. The equation of directrix is $y=-\frac{113}{16}$.

See the step by step solution

## Step 1. Write down the given information.

The given equation is $y=4{\left(x-3\right)}^{2}-7$.

## Step 2. Concept used.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum. ## Step 3. Calculation.

From the concept stated above, the parabola will be of the form $x=a{\left(y-k\right)}^{2}+h$ where $a<0$. Therefore, the equation of the parabola is written as:

$\begin{array}{c}x=a{\left(y-k\right)}^{2}+h....\left(\begin{array}{l}\text{where}a<0\text{and}\\ \left(h,k\right)\in R\text{}\left(\text{Real}\right)\end{array}\right)\\ x=-3{\left(y-3\right)}^{2}+4....\left(\begin{array}{l}\text{where}a=-3\text{and}\\ h=4,k=-3\end{array}\right)\end{array}$

## Step 4. Conclusion.

The equation of the parabola which will be open to left is $x=-3{\left(y-3\right)}^{2}+4$. ### Want to see more solutions like these? 