How do you find the vertex and the intercepts for f(x)=x^2+x-2?

2 Answers
Mar 26, 2016

Vertex (-1/2, -9/4)

Explanation:

x-coordinate of vertex:
x = -b/(2a) = - 1/2
y-coordinate of vertex:
y(-1/2) = 1/4 - 1/2 - 2 = -1/4 - 2 = -9/4
Vertex (-1/2, -9/4).
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make t = 0, and solve the quadratic equation:
x^2 + x - 2 = 0.
Since a + b + c = 0, use shortcut. The 2 x-intercepts (real roots) are:
1 and c/a = -2.

Mar 26, 2016

v(-1/2 , -9/4)

x int: (-2,0) and (1,0)

y int: (0,-2)

Explanation:

In a quadratic equation like ax^2+bx+c , the formula -b/(2a) gives us the x coordinate of the vertex. Then we put the x value we found in the equation to get the y value.

So, x=-1/2 -> f(x)=y=1/4-1/2-2 -> y=-9/4 -> v(-1/2, -9/4)

To find the x intercepts, we should value y as 0 in the equation;

0=x^2+x-2
0=(x+2)(x-1)
x=-2, x=1

To find the y intercept, we should value x as 0 in the equation;

y=-2