SOLUTION: Can you please help me with this, we are doing differential Calculus.
Find the equation of the tangetn for each given x-value.
y=x^2+8x+12, where x = -4.
At the back of the book
Question 25761: Can you please help me with this, we are doing differential Calculus.
Find the equation of the tangetn for each given x-value.
y=x^2+8x+12, where x = -4.
At the back of the book, it says the answer is y=-4
I have no idea how to get that.
thank you. Found 2 solutions by venugopalramana, Earlsdon:Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! Can you please help me with this, we are doing differential Calculus.
Find the equation of the tangetn for each given x-value.
y=x^2+8x+12, where x = -4.
SLOPE OF THE TANGENT AT A POINT IS GIVEN BY THE DERIVATIVE OF THE FUNCTION AT THAT POINT.HENCE DIFFERENTIATING THE GIVEN FUNCTION W.R.T. X ,WE GET
DY/DX=2X+8
AT X = -4 ITS VALUE IS
(DY/DX)AT X=-4 IS SLOPE OF TANGENT =2*-4+8=8-8=0
SLOPE IS ZERO MEANS THE LINE IS PARALLEL TO X AXIS
HENCE THE EQUATION IS Y= CONSTANT , WHERE CONSTANT IS TO BE FOUND USING THE FACT THE VALUE OF Y AT THE GIVEN POINT X=-4 SHOULD BE EQUAL TO THAT
HENCE Y = (-4)^2+8*(-4)+12=16-32+12=-4
HENCE Y=-4 IS THE EQUATION OF TANGENT AT X=-4
You can put this solution on YOUR website! As you probably know, the slope of the tangent at the given point (x = -4) is given by the value of the 1st derivative of the function at that point (x = -4). Substitute x = -4 and solve. So, the slope of the tangent is 0 which means that it's a horizontal line. To find the point (x, y) of tangency, you'll substitute x = -4 into the original equation.
The point of tangency is (-4, -4) and the equation of the line passing through that point with a slope of m = 0 is:
Using the point-slope form: Simplify. This is the equation of the tangent at x = -4
Here's what it looks like in a graph:
(