# Math Help - Tangent -> Slop intercept form

1. ## Tangent -> Slop intercept form

I'm struggling with the following. I have a question such as:

Find an equation for the line tangent to $y=3-5x^2$ at (5,-122)

I know how to calculate the slope which is:

$-50=(y-(-122))/(x-(5))$

And the above in slope intercept form is:

$-50x+128$

How was the slope intercept form calculated? I can't seem to figure this out.

2. I got it. The formula is y=mx+b

In this case,

y=-50x+b
-122=-50(5)+b
-122=-250+b
-122+250=b
128=b
y=-50x+128