Question

How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of `y = x^2 + 6x + 13`?

Answer 1

This is a quadratic function of form `y=ax^2+bx+c`.
In this case, a = 1, b = 6 and c = 13.

Since `a>0 =>` the arms go up.

The y-intercept value is `c=13`

The x-intercepts are when y = 0 and hence are the roots of the equation `x^2+6x+13=0`
Solving this equation we get
`x=(-b+-sqrt(b^2-4ac))/(2a)=-3+-2i in CC`, and so there is no x-intercepts.

The turning point (minimum value in this case since arms go up) is the point where the derivative of the function is zero, ie when
`2x+6=0=>x=-3`
Hence the axis of symmetry is `x=-3`
The corresponding y-value is then `y(-3)=4`
Hence the vertex occurs at `(-3;4)`

The domain is all the possible x-values allowed as inputs and in this case is all real numbers `RR`
The range is all possible y-values allowed as outputs and in this case is all real numbers greater than or equal to 4, that is the interval `[4;oo)`

Graphically :

graph{x^2+6x+13 [-16.82, 15.2, -0.92, 15.1]}