Find the equation the graphs given below.

The graph shows a parabola opening downwards. The axis of symmetry is indicated by the dotted line $x=2$. One x-intercept is clearly marked at $x=3$.
Since the axis of symmetry is $x=2$ and one x-intercept is $x=3$, the other x-intercept must be symmetric with respect to $x=2$. The distance from 2 to 3 is 1, so the other x-intercept is $2-1=1$.
(Note: The label $-3$ on the x-axis is inconsistent with the axis of symmetry $x=2$ and x-intercept $x=3$. We will proceed with $x=1$ and $x=3$ as the x-intercepts.)
The general form of a quadratic equation with x-intercepts $x_1$ and $x_2$ is $y = a(x-x_1)(x-x_2)$.

Using $x_1=1$ and $x_2=3$:

The graph shows the parabola passing through the y-axis at a negative value. Assuming the y-intercept is $(0,-3)$ (which is visually plausible and a common value for $a$ to be an integer), substitute this point into the equation:

The equation for graph a) is $\boxed{y = -x^2 + 4x - 3}$.

The graph shows a parabola opening upwards. The vertex is clearly marked at $(-1, -4)$. One x-intercept is marked at $x=1$.
The general form of a quadratic equation with vertex $(h,k)$ is $y = a(x-h)^2 + k$.

Using $h=-1$ and $k=-4$:

The graph shows an x-intercept at $x=1$, so the point $(1,0)$ is on the parabola. Substitute this point into the equation:

The equation for graph b) is $\boxed{y = x^2 + 2x - 3}$.