Let f x x2 Find the equation of the line tangent to the cur

Let f (x) = x2

Find the equation of the line tangent to the curve and going through the point (5, 1)

The Tangent line is suppose to touch the parabola at the origin and cross through the point (5,1)

so the Answer is:

How do I get that answer, please walk through step by step.

Points on the curve are (x, x²)

The slope there is 2x.

Then, the line that goes through this points and (5, 1) will have slope 2x.

The slope from (5, 1) to (x, x²) will have slope

(x² - 1)/(x-5)

Thus, (x² - 1)/(x-5) = 2x

(x-5)2x = (x² - 1)

2x² - 10x = x² - 1

x² - 10x + 1 = 0

From the quadratic formula,

x = (10 ± (10² - 4(1)(1)) )/2

x = (10 ± 96 )/2

x = (10 ± 46 )/2

x = 5 + 26

x = 5 - 26