Graphing \(y= x^2 +4\)
Step by step guide on how to graph this quadratic equation.
These numbers \( a, b, c\) will be useful for the next few steps.
\(a\) is the number before \(x^2\); \(a=1\)
\(b\) is the number before \(x\). In our case, \(b =0\) because we do not have a \(bx\) term.
\(c\) is the number that does not have any variable attached to it. \(c=4\)
We find the \(x\) coordinate of the vertex by using this formula:
\(x = \frac{-b}{2a}\)
In our case, since \(b =0\), then \(x=0\).
Tip: To save time in other cases, whenever you see \(b=0\), remember that \(x=0\) for the vertex of the parabola.
To find the \(y\) coordinate of the vertex, we substitute this value in the equation and we get:
\(x^2 + 4=0^2+4= 4\)
So the vertex is the point (\(0, 4\)) and the axis of symetry is the y-axis itself. This also means that we have found the y-intercept of this parabola, which is the vertex. Now we just need to add a few points to the left and right of the vertex. Easier than 'regular' cases, right?
Let's add (\(0, 4\)) to the graph.
Since we know that the vertex is at (\(0, 4\)) and this parabola goes up (remember that when \(a\) is positive, the parabola goes up), it will not cross the x-axis, so there are no x-intercepts. We need to find a few points to the left and a few points to the right of the vertex (\(x=0)\). We have to choose an \(x\) smaller than zero, and one bigger than zero.
\(x=-2\) , \(y=(-2)^2 + 4 =8\)
So the point (\(-2, 8\)) is part of the parabola, and so is the point (\(2, 8\)) because the y-axis is the simmetry axis.
Now, let's choose another \(x\) that is bigger than zero.
\(x=3\) , \(y=3^2 + 4 = 13\)
So the point (\(3, 13\)) is part of the parabola, and so is the point (\(-3, 13\)).
Let's add all these points to the graph.
Now that we have enough dots, we can connect them to build our parabola. Remeber that the parabola is a curve, NOT a straight line.