Thread: interesting derivative question

so i have a multipart question. the first is find the derivative of ax^2+bx+c, which i got to be 2ax+b.

the next part of the question reads the graph below is the graph of a function of the form f(x)=ax^2+bx+c. determine the sign of a, b, and c: positive , negative, or zero using the above derivative information. Answers without an explanation including the derivative will receive no credit.

the graph has a parabola opening up with intercepts at -2, 2. and a y int -4

so the c of the function is -4, i know that for sure, it was given that c was the y int of the parabola. i think that b of f prime is the y int of the tan line? so also -4? then i dont know what the a is. any help? as the question says i have to use the derivative to get the the values of a and b. this question is practice sheet is driving me nuts, and i want to see it through. ive just gotten so far (i think) id like to get it done. there are 9 other graphs on this sheet, so if someone could help me with this one, i could get the others. I just cant make the connection to what the sign of a is using the derivative.

For b, the sign relates to whether the derivative's y-intercept was positive or negative. So I would say, the graph reaches a maximum at x=0, therefore the derivative has y=0 when x=0. Therefore b=0.

For a, the sign of a relates to whether the parabola opens up or down, and also whether the line's slope is positive or negative (respectively). So I would say, the function increases on the right and decreases on the left, implying that the derivative goes from negative to positive values from left to right. Since the derivative is a line, it has constant positive slope; that is, 2a>0. So a is positive.

It seems a little artificial, but the question is rather artificial, so I'd run with it.

?but to get b dont you have to plug the 2x + b into y- y1 = 2ax + b(x - x1) ?
cause i get y- -4 = 2ax^2 + bx -4, pretty much back to the f(x)

she also gave us the hint a(x-h)^2 + k and after doing the derivative manually before i sub 0 in for the last h on top i get lim h ->0, a(2x+h) +b is b the y? can i do this, because i still have lim h->0 on the left, and that h is gonna go bye bye, does that still mean that be is k, meaning that b is the y value of my vertex?