SOLUTION: Given the quadratic equation y= -2x^2 - 16x- 33 and a line with slope -2 have just one point of intersection, what is the y-intercept of the line?

Algebra ->  Test -> SOLUTION: Given the quadratic equation y= -2x^2 - 16x- 33 and a line with slope -2 have just one point of intersection, what is the y-intercept of the line?      Log On


   

Question 1049515: Given the quadratic equation y= -2x^2 - 16x- 33 and a line with slope -2 have just one point of intersection, what is the y-intercept of the line?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Line y = highlight_green%28-2%29x+%2B+b
tangent to:
f(x)= -2x^2 - 16x- 33
f' = -4x - 16 = slope of tangent line
-2 = -4x - 16
x = -7/2
f(-7/2) = -3/2
P(-7/2, -3/2) Point of Intersection
***Using point-slope form, y+-+y%5B1%5D+=+highlight_green%28m%29%28x+-+x%5B1%5D%29
y + 3/2= -2(x+7/2)
y = -2x -17/2