Find the parabola, given two points and the slope at a point?

Question

Passes through (-4, -3) and (0,4), and the slope at x=0 is 1. Is this even possible?

The parabola should be a vertical one, an actual function.

Answer 1

y = ax² + bx + c is the general equation.

slope at any point is dy/dx = 2ax + b. at x = 0, 2a(0) + b = 1, so b = 1.

plugging in (0,4), 4 = a(0)² + 1(0) + c, so c = 4.

plugging in (-4,-3), -3 = a(-4)² + 1(-4) + 4, so -3 = 16a - 4 + 4, a = 3/16.

y = (3/16)x² + x + 4

Answer 2

y =ax^2 +bx+c
-3=16a -4b+c (point(-4,-3)
4=c point(0,4) and
the slope is y´=2ax+b which at x=0 is b=1
so -3=16a-4+4
and a= -3/16
so
y=-3/16x^2+x+4

Answer 3

Differentiating the two facets: y' = 2x/9 Slope of tangent line at (-4, sixteen/9) = 2/9 * (-4) = -8/9 Eqn of tangent line: y - sixteen/9 = -8/9 * (x + 4) y = -8x/9 - 32/9 + sixteen/9 y = -8x/9 - sixteen/9

Answer 4

so all you have is

y'(0)=1 and it passes through (-4,-3) and (0,4).

how do you know it is vertical? lol you stupped me. let me know if you get an answer.