For what values of a and b is the line 4x + y = b tangent to the parabola y = ax^2 when x = 3?

Answer 1

Slope of tangent line y' = 2ax.

On the other hand y = - 4x + b, y' = -4, 2ax = - 4, 2a *3 = -4, a = - 2/3

y = -2/3x^2, x = 3, y = -2/3*3^2 = - 6, (-3, -6) is tangent point, this point on given line 4x + y = b,

4*(-3) - 6 = b, b = - 18

Answer 2

a=-2/3

b=6

Hello. I’ll try to help. There may be is an easier way to find a and b but here is one way.

First, you need to find the slope of the tangent line. To get the slope, you can put 4x + y = b into slope intercept form, y = -4x + b. As the slope of a line is always the coefficient of x when the line is in slope intercept form, the slope is equal to -4.

The slope of the tangent line, -4, is equal to the derivative of the function, the parabola y =ax^2, evaluated at the x value, 3.

The derivative of y = ax^2 is equal to 2ax. When plugging 3 in you get 6a.

Setting 6a equal to -4 and solving, you get a = -2/3.

Plugging a into quadratic equation, gets the equation, y =-2/3x^2.

As the lines are tangent, they intersect at x=3. To find b, we need to find the y value of the point of intersection. Plugging 3 into y=-2/3x^2 gets y =-6.

The point of intersection is (3,-6)

Plugging this into the equation of the line, 4x + y = b gets b=6.

Hope this helps!