We are going to graph a quadratic function from Vertex Form. For the following function, let's look at the graph and see if we can find the key information that can be found directly from the equation.
Key Information
As you can see from the graph, the vertex is \((3, 2)\). If we look at the equation in vertex form we can see the vertex (I think the name gave it away)!! As you learned previously in Target A the vertex \((h, k)\) can be found very easily when the quadratic equation is in vertex form. The vertex is the key information that we can obtain directly from the equation. If you need to review finding the vertex when a quadratic equation is in vertex form click the following link: GQF Target A Quadratic Comparison.
Example 1: Graph the function: \(f(x) = (x - 2)^2 + 3\).
Solution:
Key Information
From the equation, we see the vertex: \((2, 3)\). This also tells us that our axis of symmetry is \(x = 2\). This is the first of our three points we need to graph the parabola.
Additional Point
To find another point we will substitute in a value for \(x\) in the equation and solve for the \(y\)-coordinate. You can choose any \(x\) value you would like since the domain of the function is all real numbers. But when you are choosing your \(x\) value pick one that is to the left or right of the \(x\)-coordinate of the vertex. Let's choose a value to the left of the \(x\)-coordinate of the vertex vertex, \(x = 0\). Substituting this value into the function.
\(f(0) = (0 - 2)^2 + 3)\)
\(f(0) = (-2)^2 + 3\)
\(f(0) = 4 + 3 = 7\)
The additional point is \((0, 7)\) - which happens to be the \(y\)-intercept.
Use Symmetry to Find a Third Point:
In order to find the third point we will use the symmetry. Let's graph the vertex, the axis of symmetry and the additional point first. Once you have graphed the additional point, then use symmetry about the axis of symmetry to find the third point. See below.
As you can see from the graph, the vertex is \((3, 2)\). If we look at the equation in vertex form we can see the vertex (I think the name gave it away)!! As you learned previously in Target A the vertex \((h, k)\) can be found very easily when the quadratic equation is in vertex form. The vertex is the key information that we can obtain directly from the equation. If you need to review finding the vertex when a quadratic equation is in vertex form click the following link: GQF Target A Quadratic Comparison.
Example 1: Graph the function: \(f(x) = (x - 2)^2 + 3\).
Solution:
Key Information
From the equation, we see the vertex: \((2, 3)\). This also tells us that our axis of symmetry is \(x = 2\). This is the first of our three points we need to graph the parabola.
Additional Point
To find another point we will substitute in a value for \(x\) in the equation and solve for the \(y\)-coordinate. You can choose any \(x\) value you would like since the domain of the function is all real numbers. But when you are choosing your \(x\) value pick one that is to the left or right of the \(x\)-coordinate of the vertex. Let's choose a value to the left of the \(x\)-coordinate of the vertex vertex, \(x = 0\). Substituting this value into the function.
\(f(0) = (0 - 2)^2 + 3)\)
\(f(0) = (-2)^2 + 3\)
\(f(0) = 4 + 3 = 7\)
The additional point is \((0, 7)\) - which happens to be the \(y\)-intercept.
Use Symmetry to Find a Third Point:
In order to find the third point we will use the symmetry. Let's graph the vertex, the axis of symmetry and the additional point first. Once you have graphed the additional point, then use symmetry about the axis of symmetry to find the third point. See below.
Lastly, we need the domain and range. As we stated previously the domain is all real numbers. As for the range, remember it is is determined by the vertex. Our vertex is a minimum with a \(y\)-value of \(3\) so the range is \(y \geq 3\).
Example 2: Graph the function \(f(x) = -(x - 2)^2 + 1\).
Solution:
Watch the video for the solution.
Example 2: Graph the function \(f(x) = -(x - 2)^2 + 1\).
Solution:
Watch the video for the solution.