# How do you find x-intercepts, axis of symmetry, maximum or minimum point, and y-intercept for the equation y = x^2+2x-3?

Apr 28, 2015

To find $x$ intercepts, find $x$ to make y=).
That is: solve ${x}^{2} + 2 x - 3 = 0$
$\left(x - 1\right) \left(x + 3\right) = 0$
The $x$ intercepts are $- 3$ and $1$.

Find $y$ intercepts by finding $y$ when $x = 0$.
$y = {\left(0\right)}^{2} + 2 \left(0\right) - 3 = - 3$
The $y$ intercept is $- 3$.

Find the vertex:

The vertex of $y = a {x}^{2} + b x + c$ has $x$ coordinate $- \frac{b}{2 a}$
So the graph of $y = {x}^{2} + 2 x - 3$ has vertex at $x = - \frac{2}{2 \left(1\right)} = - 1$
The axis of symmetry is the line: $x = - 1$

Because $a = 1$ is positive, the parabola opens upward, so there is a minimum value for $y$. That minimum occurs where $x = - 1$, so the minimum is: $y = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 3 = 1 - 2 - 3 = - 4$
The minimum value of $y$ is $- 1$.

I don't know what you've been taught to call the "minimum point", but I'll assume it has 2 coordinates, so the "minimum point" is the vertex.
I expect the "minimum point" to be $\left(- 1 , - 4\right)$.