Question

How do you graph `y=x^2-2x-5`?

Answer 1

SImple answer: with an online graphing calculator

Explanation:

Instead, you can also graph this by finding the roots. This is done by using the quadratic formula, ie `(-b+-sqrt(b^2 - 4ac))/(2a)`

The result of this is that the roots are at `x=-1.449` and `x=3.449`

The turning point will be at the point where the derivative is equal to zero. For this step, `y=x^2 - 2x - 5`

So `dy/dx = 2x - 2`

`dy/dx = 2x - 2 = 0`

`2x = 2`

`x = 1`

Finally, check the corresponding y-coordinate at `x=1`

`y=x^2 - 2x - 5`

`y = 1^2 - 2 - 5 = 1-7 = (-6)`

And as this is a quadratic, you just fill in the line in the usual `x^2` shape, making the line fit the points we have found above.

Answer 2

You can find the `"Vertex, zeroes, y-intercept, and additional points"`.

Explanation:

You can find the roots by using the quadratic formula:

`(-b+-sqrt(b^2-4ac))/(2a)`

Here,

`a=1`

`b=-2`

`c=-5`

Plug in.

`(-(-2)+-sqrt((-2)^2-4*1*-5))/(2*1)=>`

`(2+-sqrt(24))/2`

Simplify:

`(2+-sqrt(6*4))/2`

`(2+-2sqrt(6))/2`

`1+-sqrt(6)`

The y-int of the equation `ax^2+bx+c` is `c`.

The y-int here is `-5`.

To find the vertex, turn to vertex form by completing the square:

`y=a(x-h)+k` with `(h,k)` as the vertex:

`(x^2-2x+(1)^2-(1)^2)-5`

`(x-1)^2-6`

The vertex is `(1,-6)`

Here is a graph for reference: graph{x^2-2x-5 [-9, 11, -6.6, 3.4]}