Find the parabola with equation
y = ax^2 + bxwhose tangent line at (2, 12)
has the equation y = 14x − 16
dy/dx = 2ax + b
when x = 2, slope = 14
2a(2) + b = 14
4a + b = 14 or b = 14-4a
also (2,12) lies on the parabola
12 = 4a + 2b
12 = 4a + 2(14-4a)
12 = 4a + 28 - 8a
4a = 16
a=8
then b= 14-32 = -18
so the parabola is y = 8x^2 -18x
check:
dy/dx = 16x - 18
when x=2
dy/dx = 32-18 = 14
equation:
y-12 = 14(x-2)
y = 14x - 28 + 12 = 14x -16
all is good.
To find the equation of the parabola with this tangent line, we need to determine the values of a and b in the equation y = ax^2 + bx.
We know that the tangent line at (2, 12) has the equation y = 14x - 16.
First, let's find the slope of the tangent line. The slope of a line is given by the coefficient of x in the equation. In this case, the slope is 14.
Since the tangent line is also the derivative of the function y = ax^2 + bx, we can use this derivative to find the slope of the tangent line:
dy/dx = 2ax + b
Setting this derivative equal to 14, we have:
2ax + b = 14
Since the tangent line passes through the point (2, 12), we can substitute these values into the equation:
2a(2) + b = 14
Simplifying, we have:
4a + b = 14
Now, let's find the y-coordinate of the point on the parabola at x = 2. We can substitute this value into the equation y = ax^2 + bx:
y = a(2^2) + b(2)
y = 4a + 2b
Since this point lies on the tangent line, we can set y equal to the corresponding y-value on the tangent line:
4a + 2b = 12
Now, we have a system of equations:
4a + b = 14
4a + 2b = 12
We can solve this system of equations to find the values of a and b.
Subtracting the second equation from the first equation, we have:
(4a + b) - (4a + 2b) = 14 - 12
b = 2
Substituting this value of b into the first equation, we can solve for a:
4a + 2 = 14
4a = 12
a = 3
Therefore, the equation of the parabola is:
y = 3x^2 + 2x