I can't find the equation of a line with the slope of -8 that has one point of intersection with the parabola y=x^2-4x+5

Question

Im having a lot of trouble figuring out the equation of a line (that has the slope of -8) and has only one point intersecting with a parabola y=x^2-4x +5. Im not sure how to isolate b once I combine the 2 equations

Answer 1

Since the slope is $-8$ you know that the equation of the line has the form $y=-8x+b$

It intersects with the parabola $y=x^2-4x+5$

At the intersection point(s) we know that the $y$ value of the line and the parabola are equal so let's put the two equations equal to each other

$-8x+b=x^2-4x+5$

Now solving for $x$ will tell us the $x$ point(s) of the intersection

$0=x^2+4x+(5-b)$

There is only one intersection point so this quadratic equation has only one real root, can you find out what value of $b$ gives only one real root?

Answer 2

Hugh's answer is good. To use Peter's hint:

In order to touch the parabola only once, the line must be a tangent. So when does the parabola have gradient $-8$?

$$ y = x^2 - 4x -5 $$ $$ {dy\over dx} = 2x - 4 $$ So we require $$ -8 = 2x - 4 $$ i.e $$ x = -2 $$ To complete, calculate $y$ when $x=-2$ and then fit it to the line, i.e. solve $$ y = -8x + b $$ to calculate the $y$-intercept $b$.