Find expressions for the quadratic functions whose graphs are shown.
I know how to do f(x) but I'm stuck on g(x).
First, note that the y-intercept is 1.
So $\displaystyle g(x) = ax^2 + bx + 1$.
Using the other two points, we will have two equations in two unknowns to solve simultanously.
Equation 1:
$\displaystyle 2 = a(-2)^2 + b(-2) + 1$
$\displaystyle 4a - 2b + 1 = 2$
$\displaystyle 4a - 2b = 1$.
Equation 2:
$\displaystyle -2.5 = a(1)^2 + b(1) + 1$
$\displaystyle a + b + 1 = -2.5$
$\displaystyle a + b = -1.5$.
So your system of equations is...
$\displaystyle 4a - 2b = 1$
$\displaystyle a + b = -1.5$.
Solve for a and b.
The second function has the form
$\displaystyle y = ax^2+bx+c$
The y-intercept c is 1 as shown in the picture so
$\displaystyle y = ax^2+bx+1$
Now use points (-2,2)
$\displaystyle y = ax^2+bx+1$
$\displaystyle 2 = a(-2)^2+b(-2)+1$ ...(1)
and (1,-2.5)
$\displaystyle y = ax^2+bx+1$
$\displaystyle -2.5 = a(1)^2+b(1)+1$ ...(2)
and solve the sytem
We assume that:
$\displaystyle g(x) = ax^2 + bx + c$
and the graph of the function got through 3 points, they are:
$\displaystyle (-2, 2), (0, 1), (1, -2.5)$
$\displaystyle g(-2) = 4a - 2b + c = 2;$
$\displaystyle g(0) = c = 1$
$\displaystyle g(1) = a + b +c = -2.5$
So:
$\displaystyle a = -1; b = -2.5; c = 1;$
$\displaystyle g(x) = -x^2 - 2.5x + 1$