How do you find the vertex and intercepts for f(x)=3(x-2)^2+1 f(x)=3(x2)2+1?

1 Answer
Dec 15, 2015

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)y=a(xh)2+kVertex: (h,k)

In this case:

y = 3(x-2)^2 +1 -> "Vertex: " (2,1)y=3(x2)2+1Vertex: (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 + 133(x2)2+1=3(x24x+4)+1=3x212x+13

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

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

x = (-b+- sqrt(b^2-4ac))/(2a)x=b±b24ac2a
x = (12+-sqrt(12^2 -4(3)(13)))/(2(3))x=12±1224(3)(13)2(3)
x = (12 +- sqrt(-12))/6x=12±126

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