Find the parabola with equation y = ax^2 + bx whose tangent line at (1, −1) has equation y = 2x − 3.

I am completely lost on how to do this problem

y'(x) = 2ax + b
y'(1) = 2a+b = 2

y(1) = -1 = a+b

a+b = -1
2a+b = 2

2a - a + b - b = 2 - (-1)
a = 3
3+b = -1
b = -4

The parabola is:
y = 3x² - 4x

I'll assume you know basic calculus.

The tangent line represents the slope of the parabola at that point. So the derivative of the parabola is 2 (since y=2x-3 has slope 2) at (1, -1).

For the parabola, dy/dx = 2ax + b. At (1,-1), dy/dx = 2. So 2 = 2a+b.

Substitute (1,-1) into the parabola to get you -1 = a+b

Simultaneous equations here. Multiply -1 = a+b by 2 to get you -2 = 2a+2b. Subtract that with 2=2a+b to get you b= -4. Substitute b= -4 back into -1 = a+b to get you a = 3.

So the parabola is y=3x^2 - 4x

Two things to keep in mind here:

At the point 1, dy/dx = 2

Also, y(1) = -1.

dy/dx = 2ax + b

2a + b = 2

a + b = -1

a = 3

b = -4

y = 3x² - 4x

y=ax^2 + bx
dy/dx = 2ax
slope is 2 at x = 1, so 2a = 2, a = 1
y = -1 at x = 1 so 2 + b = -1, b = -3
the eqn is y = x^2 - 3x