How do you find the vertex and intercepts for $y = x^2 +2x - 5$ ?

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Answer 1 · 2015-11-24T22:54:51

See explanation.

Finding the vertex:
Use the formula for the point of the vertex $(-b/(2a),f(-b/(2a)))$ .

Remember that the quadratic is in the general form $ax^2+bx+c$ , so $a=1$ and $b=2$ .

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

$f(-b/(2a))=f(-1)=(-1)^2+2(-1)-5=-6$

The vertex is at the point $(-1,-6)$ .

Finding the y-intercept:

Plug in $0$ for $x$ .

$y=(0)^2+2(0)-5=-5$

The $y$ -intercept is $(0,-5)$ .

Finding the x-intercept:

Plug in $0$ for $y$ .

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

$x=(-2+-sqrt(2^2-4(1)(-5)))/(2(1))$

$x=(-2+-sqrt24)/2$

$x=(-2+-2sqrt6)/2=-1+-sqrt6$

The $x$ -intercepts are $(-1+sqrt6,0)$ and $(-1-sqrt6,0)$ .

How do you find the vertex and intercepts for y = x^2 +2x - 5y = x^2 +2x - 5?