Question #312318

Find the equation of the tangent to at the point where , in f(x)=x^2+4x-5

standard form.


1
Expert's answer
2022-03-16T17:46:22-0400

We proceed to differentiate: f'(x)=2x+4


Then we substitute for the point x0:


yy0=f(x0)(xx0)yy0=(2x0+4)(xx0)y=y0+(2x)x02x02+4x4x0y=x02+4x05+(2x)x02x02+4x4x0y=2x(x0+2)x025=mx+by-y_0=f'(x_0)*(x-x_0) \\ y-y_0=(2x_0+4)*(x-x_0) \\ y=y_0+(2x)x_0-2x_0^2+4x-4x_0 \\ y=x_0^2+4x_0-5+(2x)x_0-2x_0^2+4x-4x_0 \\ y=2x(x_0+2)-x_0^2-5=mx+b


This leads us to the conclusion that the standard form for the tangent line is the line with slope m=2(x0+2)m=2(x_0+2) and the intersection at b=(x02+5)b=-(x_0^2+5), where x0 is the point where the tangent line is placed.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS