How do you graph the parabola $f(x)=x^2-2x-15$ using vertex, intercepts and additional points?

Question

11516 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-06T02:28:40

Refer Explanation section

Given -

$y=x^2-2x-15$

It can be graphed by plotting the following points

vertex, y-intercept, and x-intercepts

Vertex

$x=(-b)/2a)=(-(-2))/(2 xx1)=1$

At $x=1$

$y=1^2-2(1)-15=1-2-15=-16$

Vertex $(1,-16)$

Y-Intercept

At $x=0$

$y=0^2-2(0)-15=-15$

Y-Intercept $(0, -15)$

Its x-intercepts are

At $y=0$

$x^2-2x-15=0$
$x^2-5x+3x-15=0$

$x(x-5)+3(x-5)=0$
$(x+3)(x-5)=0$
$x=-3$
$x=5$

The two x- intercepts are $(-3, 0) (5, 0)$

Plot the points $(1,-16); (0, -15); (-3, 0) (5, 0)$

You will get the curve

Look at the graph

How do you graph the parabola f(x)=x^2-2x-15f(x)=x^2-2x-15 using vertex, intercepts and additional points?