Question

Find the parabola with equation y = ax^2 + bx whose tangent line at (2, 4) has equation y = 8x − 12.

Answer 1

$\bf y=ax^2+bx\implies \cfrac{dy}{dx}=2ax+b$

now, we know the tangent line at (2,4) is y = 8x - 12, now, that's already in slope-intercept form, so, the slope of that tangent at (2,4) is "8" then.

now, that means when x = 2, the slope is 8, so let's use that.

$\bf \left. \cfrac{dy}{dx} \right|_{x=2}\implies 2a(2)+b=\stackrel{\textit{from tangent equation}}{8}\implies 4a+b=8 \\\\\\ 4a=8-b\implies a=\cfrac{8-b}{4}\implies \boxed{a=2-\cfrac{b}{4}}\\\\ -------------------------------\\\\ \begin{cases} x=2\\ y=4\\\\ a=2-\cfrac{b}{4} \end{cases}\implies \stackrel{y = ax^2+bx}{4=\left( \boxed{2-\frac{b}{4}} \right)(2)^2+b(2)}\implies 4=8-b+2b \\\\\\ \boxed{-4=b}\qquad thus\qquad a=2-\cfrac{-4}{4}\implies \boxed{a=3}\\\\ -------------------------------\\\\ y=3x^2-4x$

Answer 2

We want to get a quadratic equation by knowing its tangent line at a point. Remember that the tangent line is given by the derivative of the equation.

We will find that the parabola is:

y = 3*x^2  - 4*x

We start with the tangent line:

y₂ = 8*x - 12

The slope of this tangent line is equal to the derivative of our parabola evaluated in x = 2.

Our parabola is y = a*x^2 + b*x

Derivating we get:

y' = 2*a*x + b

Then we have:

2*a*2 + b = 8

And we also know that our parabola passes through the point (2, 4), then we must have:

4 = a*2^2 + b*2

Then we have a system of equations:

2*a*2 + b = 8

4 = a*2^2 + b*2

First we can simplify the equations:

4*a + b = 8

4 = 4*a + 2*b

We can isolate 4*a in the first equation to get:

4*a = 8 - b

Now we can replace it in the second equation:

4 = (8 - b) +  2*b

4 = 8 + b

4 - 8 = b

-4 = b

Now that we know the value of b, we can use

4*a  = 8 - b

4*a = 8 + 4 = 12

a = 12/4 = 3

Then the equation of the parabola is:

y = 3*x^2  - 4*x

If you want to learn more, you can read:

https://brainly.com/question/4074088