y'(-5)= 8(-5)-3=-43
now for the line
y= mx+b
where x,y is a point on the curve
x=-5, so y= 4x^2-3x+5 solve for the y.
Now that you have x,y, put it in the equation y=mx+b and solve for b.
The slope of the tangent line to the parabola y=4x2–3x+5 at the point where x=–5 is:______
The equation of this tangent line can be written in the form y=mx+b where m is:_________
and where b is:______?
The first to blanks are -43, but what is b? thanks
1 answer