I was given the graphs:
I was told to say whether the a, b and c coefficients in y=ax^2+bx+c were positive or negative for each graph. The a and c coefficients can be found just by looking at the graph and for g(x) it is easy to see that the a coefficient is negative and for f(x) its positive, and they are both negative for the c coefficient.
However when it came to finding the b coefficient I became a little unsure. I know that $\frac {-b}{2a}$ gives you the x value where the axis of symmetry is and my teacher told me that $\frac {-b}{2a}$ > 0 and $\frac {b}{2a}$ < 0 so I assumed that g(x)'s b coefficient would be positive since the x point (approx -2) at the axis of symmetry is negetive. And for the f(x) graph I assumed that b would be negative as the axis of symmetry is positive.
so to test this I tried to recreate all the coefficients to make similar graphs so for f(x) I had these coefficients: a= 1 c = -2 and I found b using $\frac {-b}{2*1}$ = 3
-b = 6
b = -6
Upon plotting the equation x^2-6x-2 (-12/12) I got a very similar looking graph.
However when it came to doing g(x) I ran into some problems. I used the coefficients a = -2 and c = -7 (anything around this range: 7-9). I was finding b under the assumption that
$\frac {b}{2a}$ <0
so:
$\frac {b}{2*-2} = -2
b = -2*-4
b = 8.
so I checked my function: -2x^2+8x-9 with an online plotter and the graph was in the positive end rather than the negative end as shown in my image. I was only able to recreate the same graph with b as a negative, but I dont understand how that would be possible since I found b to be positive in my working. What am I doing wrong here?
For $y=g(x)$, we need to find a $b$ such that $\frac{-b}{2a}<0$. This only happens if $b<0$ since $a<0$. For instance, let $a=-1$ and $b=-2$. Then $\frac{-b}{2a}=\frac{-(-2)}{2(-1)}=-1<0$ You can see that if $b>0$, then our ratio wouldn't be negative. I'll leave it up to you to find $c$.