The parabola shown is the graph of f(x) = Ax^{2} + 2x + C. The x-intercepts of the graph are at -4 and -3. Find the exact value of the y-intercept and the coordinates of the vertex of the graph (expressed in terms of rational numbers and radicals).

Given:
$f(x)=A{x}^2+2x+C$
x intercept are $-4and-3$
We need to write x intercept in factor form
If p and q are the factors then intercepts form of quadratic equation is
$f(x)=a(x-p)(x-q)$
$-4and-3$ are x intercepts
$f(x)=a(x-(-4))(x-(-3))$
$f(x)=a(x+4)(x+3)$
$f(x)=a(x^2+7x+12)$
$f(x)=a{x}^2+7ax+12a$
Now compare f(x) with out given f(x)
$f(x)=A{x}^2+2x+C$
$f(x)=a{x}^2+7ax+12a$
$7a=2$
$a=\frac{2}{7}$
Replace the value in f(x), then f(x) becomes
$f(x)=a{x}^2+7ax+12a$
$f(x)=\frac{2}{7}{x}^2+7(\frac{2}{7})x+12(\frac{2}{7})$
$f(x)=\frac{2}{7}{x}^2+2x+\frac{24}{7}$
To find out vertex we use formula
$x=\frac{-b}{2a}$
$b=2,a=\frac{2}{7}$
$x=\frac{-2}{2\left(\frac{2}{7}\right)}=\frac{-2}{\frac{4}{7}}=-2\cdot \frac{7}{4}=\frac{-7}{2}$
now we find out y
$f(x)=\frac{2}{7}{x}^2+2x+\frac{24}{7}$
$x=\frac{-7}{2}$
$f(x)=\frac{2}{7}{\left(\frac{-7}{2}\right)}^2+2\left(\frac{-7}{2}\right)+\frac{24}{7}=\frac{-1}{14}$
Vertex
$(\frac{-7}{2}\cdot \frac{-1}{14})$
Now find out y intercept when $x=0$
$f(x)=\frac{2}{7}{x}^2+2x+\frac{24}{7}$
$f(0)=\frac{2}{7}(0{)}^2+2(0)+\frac{24}{7}=\frac{24}{7}$
y intercept is $(\frac{24}{7})$