Find an equation of the tangent line to the graph of the function f(x)=x^2+3x+5 at point (-4,9)

Answer 1

Hi Celine, To find an equation of the tangent line I always start in point - slope form:

y - y₁ = m (x - x₁)

We are already given the point (-4, 9), so we will plug those values in for x₁ and y₁ in the above equation.

To find the slope, m, of the tangent line, we need to find the derivative of the given function and plug in x = -4:

f'(x) = 2x + 3

f'(-4) = 2(-4) + 3 = -5, so m = -5.

Now, write the equation in point-slope form:

y - 9 = -5 (x + 4)

If you need to write the equation in slope-intercept form, solve for y:

y = -5x -11

In standard form, the equation will be: 5x + y = -11

I hope this helps. Please let me know if you have any more questions.

Answer 2

The derivative of a function evaluated at a point is the slope of the tangent line to the curve at that point.

df/dx = 2x +3 using power law and that constants have a derivative of zero.

df/dx(x = -4) = 2(-4) + 3 = -5

This is the slope of the tangent line

You can put it in the form y = mx + b input x = -4, y = 9, and m = -5 --> solve for b or

You can rearrange the definition of slope between an arbitrary point on the line (x,y) and the point you know (-4, 9):

(y-9)/(x+4) = -5 solve for y = ...

Hope that helps.

Answer 3

Hey! Lets solve this shall we ?

So, we must find the equation of a tangent line given a function and point. First, we take the first derivative of f using the power rule such that

f '(x) = 2x + 3

Then, we evaluate using x = -4 of the given point (-4,9) and plug it into the derivative such that

f ' (-4) = 2(-4) + 3 = -5

Next we use the point-slope formula to find the equation of a tangent line such that

y - y1 = f ' (-4) ( x - x1)

y - 9 = (-5)(x - (-4))

y - 9 = -5(x + 4)

y - 9 = -5x - 20...........Then add 9 such that

y = -5x - 11

I hope this helped!