Calculus Homework

Question

(a) Find the slope m of the tangent to the curve y = 9 + 4x^2 − 2x^3 at the point where x = a.

m=8a-6a^2

(b) Find equations of the tangent lines at the points (1, 11) and (2, 9).

y(x)=_______ (at the point 1,11)

y(x)=_______ (at the point (2,9)

Answer 1

Part (a) is correct.

For part (b), any time you need "tangent line" think of the pt slope equation: y-y1=m(x-x1)

Remember a derivative is a slope.

For point (1,11) you need to find m(1), which would be 2. Hence, y-11=2(x-1)

For point (2,9) you need m(2) which would be -8. So it would be: y-9=-8(x-2)

I hope this helps!

Answer 2

(a) Find the slope m of the tangent to the curve y = 9 + 4x² − 2x³ at the point where x = a.

You already have m = 8a – 6a² which is correct.

b) Find equations of the tangent lines at the points (1, 11) and (2, 9).

tangent line at (1, 11)

m = 8(1) – 6(1)² = 2

You just need the y-intercept (b) and you have the equation for the tangent line y = 2x + b.

y = 2x + b means that b = y – 2x

To find b you can plug in x = 1 and y = 11

b = 11 – 2(1) = 9

So the equation for the tangent line is y = 2x + 9

tangent line at (2, 9)

m = 8(2) – 6(2)² = –8

So we have y = –8x +b

To find b you can plug in x = 2 and y = 9 into b = y + 8x

b = 9 + 8(2) = 25

So the equation for the tangent line is y = –8x + 25