Question

How do you graph the parabola `y=x^2 + 3x-10` using vertex, intercepts and additional points?

Answer 1

Graph the y-intercept (0,-10), the x-intercepts (-5,0) and (2,0) and the vertex `(-3/2,-49/4)`

Explanation:

To find the y-intercept, let x = 0 in the original equation:
`y=x^2+3x-10`
`y=(0)^2+3(0)-10`
`y=-10`
So the y-intercept is (0, -10)

To find the x-intercepts, let y=0
`y=x^2+3x-10`
`0=x^2+3x-10`

Now, factor the expression:
`0=(x+5)(x-2)`

Using the zero product rule, set each bracket equal to zero:
`x+5=0` or `x-2=0`
Giving the result
`x=-5` and `x=2`
So the x-intercepts are `(-5,0) and (2,0)`

The vertex is on the line (called the axis of symmetry) that is half-way between the x-intercepts. To find the half-way value, find the midpoint between the x-intercepts:
`(x_1+x_2)/2=(-5+2)/2`
`color(white)(aaaaaaa)` `=-3/2`
So the axis of symmetry is `x=-3/2`, and to find the y value of the vertex, substitute this value for x into the equation `y=x^2+3x-10`
`y=(-3/2)^2+3(-3/2)-10`
`y=9/4-9/2-10`
Getting common denominators and adding,
`y=9/4-18/4-40/4`
`y=-49/4`
Therefore the vertex of the parabola is `(-3/2,-49/4)`

Graph the parabola by plotting the vertex, the y-intercept and the x-intercepts and an additional point (-3,-10) which is the matching point to the y-intercept on the other side of the axis of symmetry. These 5 points should give a reasonable representation of the graph of the parabola.