# How do you graph y = x^2 + 7x - 1?

The given equation can be changed into a standard equation of parabola: ${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$ having vertex at $\left({x}_{1} , {y}_{1}\right)$ & focus at $\left({x}_{1} , {y}_{1} + a\right)$

#### Explanation:

Given equation: $y = {x}^{2} + 7 x - 1$
$y = {x}^{2} + 2 \setminus \cdot \setminus \frac{7}{2} \setminus \cdot x + {\left(\setminus \frac{7}{2}\right)}^{2} - {\left(\setminus \frac{7}{2}\right)}^{2} - 1$
$y = {\left(x + \setminus \frac{7}{2}\right)}^{2} - \setminus \frac{51}{2}$
${\left(x + \setminus \frac{7}{2}\right)}^{2} = y + \setminus \frac{51}{2}$
${\left(x + \setminus \frac{7}{2}\right)}^{2} = 4 \setminus \cdot \setminus \frac{1}{4} \left(y + \setminus \frac{51}{2}\right)$
comparing the above equation with ${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$
we get ${x}_{1} = - \frac{7}{2} , {y}_{1} - \frac{51}{2} , a = \frac{1}{4}$
then the given parabola has vertex at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- \frac{7}{2} , - \frac{51}{2}\right)$ & focus at $\left({x}_{1} , {y}_{1} + a\right) \setminus \equiv \left(- \frac{7}{2} , - \frac{101}{4}\right)$
Points of intersection of parabola: $y = {x}^{2} + 7 x - 1$ with x-axis where $y = 0$ are $\left(\setminus \frac{- 7 + \setminus \sqrt{53}}{2} , 0\right)$ & $\left(\setminus \frac{- 7 - \setminus \sqrt{53}}{2} , 0\right)$ and the point of intersection with y-axis is $\left(0 , - 1\right)$
Now, locate the vertex & locus & draw the axis of parabola. Draw free hand curve passing through above points of intersection with the coordinate axes