If f(x) = 7 + 2x - 5x^2, find the equation of the line tangent to the curve f( x ) at x=-5?

f(x) = 7 + 2x - 5x^(2)

Given that x = - 5, plug that into the f(x) function here to determine the y - value.

f( - 5 ) = y = 7 + 2 ( - 5 ) - 5 ( - 5 )^(2) = 7 - 10 - 125 = - 128

Therefore y = - 128.

Take the derivative with respect to both sides of the function with respect to x.

d/dx[f(x)] = d/dx[7 + 2x - 5x^(2)]

d/dx[f(x)] = d/dx[7] + d/dx[2x] + d/dx[ - 5x^(2) ]

f ' (x) = 2 - 10x

Factor out a 2 from the derivative function.

f ' (x) = 2 [ 1 - 5x ]

Given that x = - 5, plug that into the derivative function to determine the slope (m) of the tangent line.

m = f ' ( - 5 ) = 2 [ 1 - 5( - 5) ] = 2 [ 1 + 25 ] = 2 [ 26 ] = 52

So since x1 = - 5, y1 = - 128, and m = 52 use the point - slope formula:

y - y1 = m (x - x1)

Substitute these quantities here.

y - (-128) = 52 (x - (-5) )

Double negatives make positives.

y + 128 = 52 (x + 5)

Multiply using the distributive property.

y + 128 = 52x + 260

Subtract 128 to both sides.

y = 52x + 260 - 128

y = 52x + 132 -------------> EQUATION OF THE TANGENT LINE ----> ANSWER

So first you want to plug in -5 for x to get your y value. And then you want to find the slope which will be the "m" that goes into the equation y=mx+b of your tangent line. But to do that, you need to use the power rule to get your derivative. And all that is is that for any function x^n, it's derivative is nx^(n-1). And that gets multiplied by any coefficients. So for 7x^0, (the x^0 is implied) you multiply that 0 by everything and the whole term goes away. For your 2x^1 (again, the power is implied) it's 2*1x^(1-1), or just 2. Then for your -5x^2, it's (-5*2)x^(2-1), or -10x. So your slope is going to be, at any point x, 2-10x. And since your x is -5, plug that in to get 2-10*(-5), or 52. Plug that in for m, -5 in for x, and the y you got earlier in for y to get your b value. The equation of your tangent line should be y=mx+b, but with the m and b you found!

We know the slope of the line will have a value equal to the derivative at the point"

g(x) = d f(x)/dx = 2 - 10x

g(5) = 52

then we find the value of f at -5:

f(-5) = -128

then we plug into the equation y = mx + b:

-128 = (52)(-5) + b

b = 132

y = 52x + 132

f(x) = 7 + 2x - 5x²

f(-5) = 7 + 2(-5) - 5(25) = -128

f'(x) =2 - 10x

f'(-5) = 2+50 = 52

Tangent line has slope 52 through point (-5, -128)

y+128 = 52(x + 5)

y = 52x + 132