# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of y=x^2+2x-3?

May 22, 2017

"vertex":(-1, -4); " "x"-intercepts": (1, 0), (-3, 0);
$y \text{-intercept":(0, -3); " Domain: all reals; Range: } \left[- 4 , \infty\right)$

#### Explanation:

Given: $y = {x}^{2} + 2 x - 3$

Find the vertex : When $y = A {x}^{2} + B x + C = 0$ the vertex is $\left(h , k\right)$, where $h = \frac{- B}{2 A}$.

$h = \frac{- B}{2 A} = - \frac{2}{2 \cdot 1} = - 1$

$k = f \left(h\right) = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 3 = - 4$

$\text{vertex} : \left(- 1 , - 4\right)$

Find $x$-intercepts by setting $y = 0$ and by factoring: $\text{ } y = \left(x - 1\right) \left(x + 3\right) = 0$

$x \text{-intercepts} : \left(1 , 0\right) , \left(- 3 , 0\right)$

Find the $y$-intercept by setting $x = 0$:

y = --3; " "y"-intercept":(0, -3)

Find Domain and range :
Domain is all the valid input $\left(x\right)$. Since there is no radical such as a square root, or no denominator that could become undefined, the domain is all valid Reals.

Domain: all Reals, $\text{ "-oo < x , oo " or } \left(- \infty , \infty\right)$

Range is limited by the input. The lowest $y$-value is the vertex.

Range: $\text{ "-4 <= y < oo " or }$[-4, oo)#