hepl with calc????

Question

find a function y=ax^2+bx+c whose graph has x-intercept=1, y-intercept=-2 and has a tangent line with a slope of -1at the y-intercept.
Please sho me the steps.

ans:3x^2+x-2

thanks

Answer 1

You need to find the variables a, b, and c that make this work out.

Since the y-intercept is 2, you know that the point (0,2) is on the function. So, if you plug in 0 for x, you should get 2 for y. That means 2 = a*0^2 + b*0 + c ... so 2 = c. This means that you have y = ax^2 + bx + 2.

You know the slope at the y-intercept is -1. The derivative of this function is y' = 2ax + b, so if you plug in 0 for x, 2*a*0 + b = -1. This means b = -1. So you have y = ax^2 - 1x + 2.

Finally you know the x-intercept is 1, which means the point (1,0) iks on the function. So if x = 1, then y = 0. Plugging into what we already have, this gives 0 = a*1^2 - 1*1 + 2 or 0 = a + 1. Solving this you get a = -1, so the function is y = -1x^2 - 1x + 2.

Answer 2

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