Sorry if the explanation is very long (it's actually very simple), I had to make sure people would be able to fully understand what was going on. If it was shorter, there could be lots of confusion.
The x-intercept(s) can be found by making y=0 then solving for x;
2(x+3)^2 + 6 = 0
2(x+3)^2 = -6
(x+3)^2 = -3
At this stage you'll notice that negative numbers do not have square roots, and so there is it is impossible to find a value for x (when y=0), meaning there are no x-intercepts, and the graph will not cross the x-axis at any point.
To make sure, you could check by finding the discriminant (Delta) of the line;
Delta = b^2 - 4 a c
Where a is the coefficient of x^2, b is the coefficient of x^1, and c is the constant (coefficient of x^0). Make sure you simplify the equation fully before taking the coefficients.
If Delta >0, there are two x-intercepts, and the graph is said to have distinct roots.
If Delta =0, there is one x-intercept and the graph is said to have repeating roots.
If Delta <0, there are no x-intercepts, and the graph is said to have imaginary roots.
In this case;
Delta = (12)^2-4(2)(24)
Delta = -48
Delta <0 and so x has imaginary roots. Confirming our judgment.
The y-intercept(s) can be found by making x=0 then solving for y;
y = 2(0+3)^2 + 6
y = 18+ 6
y = 24
The vertex can be found by carrying out either of two methods;
- Memorizing and using this expression:
Vertex is (-b//2a, 4ac-b^2//4a)
Where as stated above, a, b and c are the coefficients of indices of x in the simplified equation.
(I'd rather not use this method though.)
2. Completing the square, then taking the coordinate values from it.
In the equation y=a(x+p)^2+q, the vertex is ( -p,q)
So in our case, p is 3 and q is 6, which means the vertex is (-3, 6).
a will tell you if the quadratic curve is positive or negative, and so if the vertex is a maximum or minimum value for y in the curve. This is referred to as the "nature of the vertex" in exam-type questions.
If a is positive (like it is in out case), then the graph is positive and the vertex is at the minimum value for y. The graph doesn't have a maximum value for y. The reason for this is obvious if you observe the shape of a positive quadratic graph.
The equation you submitted is already in completed square format and so this is obviously the easier method of the two.
