Question

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

Answer 1

The given equation can be changed into a standard equation of parabola: `(x-x_1)^2=4a(y-y_1)` having vertex at `(x_1, y_1)` & focus at `(x_1, y_1+a)`

Explanation:

Given equation: `y=x^2+7x-1`
`y=x^2+2\cdot \frac{7}{2}\cdotx+(\frac{7}{2})^2-(\frac{7}{2})^2-1`
`y=(x+\frac{7}{2})^2-\frac{51}{2}`
`(x+\frac{7}{2})^2=y+\frac{51}{2}`
`(x+\frac{7}{2})^2=4\cdot \frac{1}{4}(y+\frac{51}{2})`
comparing the above equation with `(x-x_1)^2=4a(y-y_1)`
we get `x_1=-7/2, y_1-51/2, a=1/4`
then the given parabola has vertex at `(x_1, y_1)\equiv(-7/2, -51/2)` & focus at `(x_1, y_1+a)\equiv(-7/2,-101/4 )`
Points of intersection of parabola: `y=x^2+7x-1` with x-axis where `y=0` are `(\frac{-7+\sqrt{53}}{2}, 0)` & `(\frac{-7-\sqrt{53}}{2}, 0)` and the point of intersection with y-axis is `(0, -1)`
Now, locate the vertex & locus & draw the axis of parabola. Draw free hand curve passing through above points of intersection with the coordinate axes