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

2 Answers
Feb 21, 2018

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.

Feb 21, 2018

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]}