Question

How do you graph the parabola `y = (x − 3)(4x + 2)` using vertex, intercepts and additional points?

Answer 1

x-intercepts: `(3,0)` & `(-1/2,0)`

y-intercept: `(0, -6)`

vertex: `(5/4,49/4)` or `(1(1/4),-12(1/4))`

graph{y=4x^2-10x-6 [-23.67, 21.94, -13.23, 9.59]}

Explanation:

First, since the function is already in factored form, I would set the function = 0.

`x-3=0 and 4x+2=0`

Solve for `x` in both of those equations to get your `x`-intercepts, which are `3` and `-1/2`

Then I would multiply those terms in the original equation to get

`y=4x^2-10x-6`

Which tells us, when we plug `0` in for `x`, that the `y`-intercept is at `(0,-6)`.

Lastly, we use our new equation

`y=4x^2-10x-6`

to find the vertex, which is `-b/(2a)`. Here `b` is `-10` and `a` is `4`, so we just plug it in:

`-(\(-10)\)/(2*4) = 5/4`

Plug in `5/4` (our `x`-value for the vertex) to either form of our equation and we get `49/4`.