SOLUTION: The tangent to the curve y = 2x^2 + ax + b at the point (-2,11) is perpendicular to the line 2y = x + 7. Find the value of a and b.

Question 1190934: The tangent to the curve y = 2x^2 + ax + b at the point (-2,11) is perpendicular to the line 2y = x + 7. Find the value of a and b.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

We'll need the derivative.
y = 2x^2 + ax + b
dy/dx = 4x + a

Plug in the x coordinate of (-2,11)
dy/dx = 4x + a
dy/dx = 4(-2) + a
dy/dx = -8 + a
This is the slope of the tangent line at the point (-2,11)

Let's solve the other equation for y
2y = x+7
y = x/2+7/2
y = 0.5x+3.5
The slope is 1/2 = 0.5
The tangent slope we found earlier is perpendicular to this. The negative reciprocal of 1/2 is -2/1 or just -2.

Set the slope of -8+a equal to the -2 and solve for 'a'
-8+a = -2
a = -2+8
a = 6

We get
dy/dx = -8+a
dy/dx = -8+6
dy/dx = -2
which helps confirm things so far.

The original equation
y = 2x^2 + ax + b
updates to
y = 2x^2 + 6x + b

Now plug in (x,y) = (-2,11) and solve for b
y = 2x^2 + 6x + b
11 = 2(-2)^2 + 6(-2) + b
11 = -4 + b
b = 11+4
b = 15

Visual confirmation using a graph

Graph was made with GeoGebra.

Answers:
a = 6
b = 15

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