Find the parabola with equation y = ax2 + bx whose tangent line at (2, 8) has equation y = 12x − 16.

Step-by-step explanation:

Before we can calculate the equation of the parabola in the form

y = ax²+bx, we need to get the value of a and b.

First we need to set up a simultaneous equation as shown;

since y = ax²+bx

dy/de = y' = 2ax + b

Substitute the point (2, 8) into the equation

y' = 2a(2)+b

y' = 4a + b

Since y' is also the slope of the equation y = 12x-16, hence y' = 12 on comparison.

The equation becomes;

12 = 4a+b

4a+b = 12 1

Since the point (2, 8) is also on the parabola, then it must satisfy the equation y = ax²+bx

substitute the coordinate;

8 = a(2)²+b(2)

8 = 4a + 2b

4 = 2a + b

2a+b = 4 2

Solve equations  1 and 2 simultaneously

4a+b = 12 1

2a+b = 4 2

Substract both equation

4a - 2a = 12-4

2a = 8

a = 4

substitute a = 4 into equation 2 and get b;

From 2; 2a+b =4

2(4)+ b = 4

8+b = 4

b = 4-8

b = -4

Substitute a = 4 and b = -4 into the equation of the parabola y = ax²+bx

y = 4x²+(-4)x

y = 4x²-4x

y = 4x(x-1)

Hence the required equation of the parabola is y = 4x(x-1)