## What are the vertex focus and directrix of the parabola with the given equation y=1/28(x-4)^2-5

October 14, 2019// Blog//

To find the equation of a line that is parallel to your original equation and goes through a certain point on a graph, here’s what you need to know:

First you need to find the slope of your original equation.

To do that, you need to convert it to slope intercept form (y = mx+b).

Add the x over, and then divide everything by 5 to get the y by itself.

Here’s what that would look like (without the small steps that I mentioned):

-x + 5y = 25

5y = x + 25

y = 1/5x + 5

That’s the original equation rewritten in slope intercept form.

The m represents the slope, so this equation’s slope is 1/5.

Because you are given a point, and now you have a slope, the best and easiest route is using point slope form.

I’ve seen different versions of the equation base but I prefer y – y(sub1) = m(x – x(sub1))

But since I can’t use subscripts in this, I’ll use the one with h and k. The h is the x value of the point, and the k is the y value.

(h,k)

Then just substitute the values in and solve for y.

y – k = m(x – h)

y + 5 = 1/5(x + 5)

y + 5 = 1/5x + 1

y = 1/5x – 4

Your final answer is

y = 1/5x – 4

You can double check by using a graph. If the slopes are the same, the lines should be parallel.

I hope that helps. If anything didn’t make sense, feel free to ask me.