Question

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

Answer 1

`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`.

Answer 2

`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`