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 25762: 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.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Find the dirivative of y = x^2 + 8x + 12
It is:
y' = 2x + 8
for any value of x, y' will be the slope of the tangent at that point.
If x = -4, then
y' = (2 * (-4)) + 8
y' = -8 + 8 = 0
so, the slope is 0 when x = -4. When the slope is zero, the tangent to
the curve is parallel to the x-axis
What is y when x = -4?
y = (-4)^2 + 8(-4) + 12
y = 16 -32 + 12 = -4
The tangent with 0 slope always takes the form y = f(x)where x is the
x-coordinate at the point where the slope is 0.
So, the equation of the tangent is y = -4
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