Question

How do you find the vertex and intercepts for `f(x)=3(x-2)^2+1`?

Answer 1

Vertex: (2,1)
x-intercepts: None
y-intercept: (0,13)

Explanation:

The quadratic given is already in vertex form, so finding the vertex is relatively simple:

`y = a(x-h)^2 + k -> "Vertex: " (h,k)`

In this case:

`y = 3(x-2)^2 +1 -> "Vertex: " (2,1)`

To find the intercepts, we need to convert to standard form by evaluating the right hand side:

`3(x-2)^2 + 1 = 3(x^2 - 4x + 4) + 1 = 3x^2 -12x + 13`

The y-intercept is simply `(0,c)` where `c` is the constant term. In this case, `c=13` and the y-intercept is `(0,13)`.

We can use the quadratic formula to find the x-intercepts:

`x = (-b+- sqrt(b^2-4ac))/(2a)`
`x = (12+-sqrt(12^2 -4(3)(13)))/(2(3))`
`x = (12 +- sqrt(-12))/6`

And here we reach a problem, because the numerator contains an imaginary number (`sqrt(-12)`), meaning that there are no x-intercepts and no real roots to this quadratic (though there are still 2 complex roots).