# How do you use the important points to sketch the graph of y=x^2+6x+5?

Mar 24, 2016

The vertex V of this parabola is $\left(- 3 , - 4\right)$. Axis is x - $- 3$,. upwards in y-direction. It cuts x-axis at $A \left(\frac{1}{2} , 0\right)$and B$\left(- \frac{7}{2} , 0\right)$.

#### Explanation:

The equation can be rearranged as ${\left(x + 3\right)}^{2} = 4 \left(\frac{1}{4}\right) \left(y + 4\right)$.

This represents a parabola with vertex at $\left(- 3 , - 4\right)$.

The parameter for size a = $\frac{1}{4}$.

The curve is above this vertex and is symmetrical about its axis $x = - 3$. This is parallel to y-axis.

Put y= o.and solve ${x}^{2} + 6 x + 5 = 0$

$x = - \frac{1}{2} \mathmr{and} - \frac{7}{2}$. So, the curve cuts x-axis at $\left(\frac{1}{2} , 0\right) , \left(- \frac{7}{2} , 0\right)$

The focus $\left(- 3 , - \frac{15}{4}\right)$ is right above vertex at a distance $\frac{1}{4}$.

The directrix #y--17/4) is equidistant below the vertex and is parallel to the x-axis..

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