Find the slope of the tangent line to the parabola y = x2 + 8x at the point (−3, −15) by using the following parameters.

Question

(a) Find the slope of the tangent line to the parabola y = x² + 8x at the point (-3, -15) by using the following parameters.

(i) The tangent line to the curve y = f(x) at the point P(a, f(a)) is the line through P with slope m=lim_h->a (f(a+h)-f(a)) / ((x-a) provided that this limit exists. m=?

(ii) The expression of slope is m=(f(a+h)-(f(a)) / h. m=?

(b) Find an equation of the tangent line in part (a).

Can someone please walk me through this? Thanks!

Answer 1

First of all, in the statement of the problem, part (i) is incorrect. x - a should be h and lim_h→a should be

lim_h→0 .

[f(a+h) - f(a)] / h = [(a+h)² + 8(a+h) - (a² + 8a)] / h = [a²+2ah + h²+8a+8h-a²-8a] / h = (2ah + h² + 8h] / h

= 2a + h + 8

Taking the limit as h approaches 0, we get 2a + 8

So, when a = -3, we have m = 2

Equation of tangent line: y - (-15) = 2[x - (-3)] So, y = 2x - 9.

Answer 2

y= x^2 +8x

y(-3) = 9-24 = -15

(-3,-15) is on the curve (worth checking before going too far in calculations)

y'= 2x +8

y'(-3) = 2(-3)+8 = 2

slope of the tangent line at (-3,-15) = 2

but you want to use the limit of the quotient difference. You should get the same answer as using differential calculus short cut rules

f(x) = x^2 + 8x

f(x+h) = (x+h)^2 + 8(x+h)= x^2 +2xh +h^2 + 8x + 8h

f(x+h)- f(x) = 2xh +h^2 +8h (the x^2 and 8x terms cancel out)

divide by h

= 2x +h +8

let h be zero

= 2x +8

the tangent line equation has slope = 2(-3)+8 =2

= (y+15)/(x+3)

y+15 = 2(x+3) is the equation in point slope form

or

y = 2x +6-15

y = 2x -9 in slope intercept form

or

2x -y = 9 in standard form