# How do you find the vertex and intercepts for y=-x^2-2x+3?

Jun 22, 2017

Vertex (-1, 4)
x-intercepts: x = 1 and x = -3

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{2}{-} 2 = - 1$
y-coordinate of vertex:
y(-1) = - 1 + 2 + 3 = 4
Vertex (-1, 4).
To find y-intercept, make x = 0 --> y = 3
To find x-intercepts, make y = 0, and solve:
$- {x}^{2} - 2 x + 3 = 0$
Since a + b + c = 0, use shortcut.
The 2 real roots (x-intercepts) are:
x = 1 and $x = \frac{c}{a} = \frac{3}{- 1} = - 3$
graph{- x^2 - 2x + 3 [-10, 10, -5, 5]}

Jun 22, 2017

The vertex is $\left(- 1 , 4\right)$, and is the maximum point of the parabola.

The x-intercepts are $\left(1 , 0\right)$ and $\left(- 3 , 0\right)$.

The y-intercept is $\left(0 , 3\right)$.

#### Explanation:

$y = - {x}^{2} - 2 x + 3$ is the equation of a parabola in standard form: $y = a {x}^{2} + b x + c$, where $a = - 1$, $b = - 2$, and $c = 3$.

The vertex of a parabola is the minimum or maximum point, $\left(x , y\right)$, of the parabola.

To determine the vertex of a parabola from the standard equation, use the following formulas:

$x = \frac{- b}{2 a}$

$y = f \left(h\right)$.

Vertex of Parabola

$x = \frac{- \left(- 2\right)}{2 \cdot - 1}$

$x = \frac{2}{- 2}$

$x = - 1$

To determine $y$, substitute the value for $x$ into the standard equation and solve.

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

$y = f \left(h\right) = - 1 + 2 + 3$

$y = f \left(h\right) = 4$

The vertex is $\left(- 1 , 4\right)$, and is the maximum point of the parabola.

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$\text{x-intercepts}$

The x-intercepts are the values for $x$ when $y = 0$. Substitute $0$ for $y$.

$- {x}^{2} - 2 x + 3 = 0$

Find two numbers that when added equal $- 2$ and when multiplied equal $3$. $1$ and $- 3$ meet the criteria.

$- \left(x - 1\right) \left(x + 3\right) = 0$

Multiply both sides by $- 1$.

$\left(x - 1\right) \left(x + 3\right) = 0$

Solutions for $x$.

$\left(x - 1\right) = 0$

$x = 1$

$\left(x + 3\right) = 0$

$x = - 3$

The x-intercepts are $\left(1 , 0\right)$ and $\left(- 3 , 0\right)$.

Determine the y-intercept by substituting $0$ for $x$ in the standard equation.

$y = - 1 \left(0\right) - 2 \left(0\right) + 3$

The y-intercept is $\left(0 , 3\right)$.

graph{y=-x^2-2x+3 [-10, 10, -5, 5]}