Question

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

Answer 1

After calculating the vertex and intercepts and plotting them, calculate additional points, `(x,f(x))`, near these crucial spots of our parabola.

Explanation:

To find the vertex, we use the formula `x = -b/(2a)`, but as you may have guessed, this will only give us the `x` coordinate of the vertex. In order to get the `y` coordinate, we would need to put our `x` value back into `f(x)`.

Standard form is
`ax^2 + bx - 10`

In our case:
`a = 1`
`b = 3`

This means the `x` coordinate of our vertex will be

`-(3)/(2(1))` or simply `-3/2`

Plugging the `x` coordinate back into `f(x)`, we get the `y` coordinate.

`(-3/2)^2+3(-3/2)-10`

`= 9/4-9/2-10`

`= -49/4`

Our vertex is therefore `(-3/2,-49/4)`.

Now, we can calculate the `x`-intercept and the `y`-intercept, which, alongside our vertex, will both help us determine what other nearby points we may wish to calculate.

The `x`-intercept is where `y=0`. This case may be solved by factoring.

`x^2 + 3x -10 = 0`

`(x + 5)(x - 2) = 0`

`x + 5 = 0 or x - 2 = 0`

`x = - 5 or x = 2`

So, our `x`-intercepts are `(-5,0)` and `(2,0)`.

The `y`-intercept is where `x=0`. Simply plug this value into `f(x)`.

`(0)^2+3(0)-10`

`= -10`

So, our `y`-intercept is `(0,-10)`.

Now, we have four relevant points that we can plot. These are:

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

After plotting these, we likely wish to plot some additional points in between to more accurately graph it. We would choose `x` values and plug them into `f(x)`.

These would be points within our range of crucial coordinates, so anywhere between our lowest `x` coordinate, `(-5,0)`, and our greatest `x` coordinate, `(2,0)`.

You may also plot more points outside of this range to give a more complete picture.

Answer 2

Explanation:

I have handwritten my explanation, I hope it is legible :)

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