# Let f(x)= x^{2}-8x+5. Find the values of x for which the slope of the curve y=f(x) is 0. The point(s) at which the slope of the tangent line is 0.

**Solution:**

Given f(x)= x^{2}-8x+5=y

To find the slope, we need to differentiate f(x)

f'(x) = 2x-8

Given that slope is equal to 0

⇒ 2x-8 =0

⇒2x=8

⇒x=4

Put x=4 in f(x), we get

f(x) = 42 -8(4) +5 = 16-32 +5 = -11

The point is (4,-11) where the slope of the tangent of f(x) will be zero.

## Let f(x)= x^{2}-8x+5. Find the values of x for which the slope of the curve y=f(x) is 0. The point(s) at which the slope of the tangent line is 0.

**Summary:**

If f(x)= x^{2}-8x+5, then the values of x for which the slope of the curve y=f(x) is 0 is 4 and the point at which the slope of the tangent line is 0 is (4,-11).