Vertex Form
In the Solving Quadratic Equations Unit, the connection between the graph of a quadratic function and its equation was introduced (click here to review). This concept focused on writing quadratic functions in both factored form and standard form based on the graph of a function. Factored form, \(y = a(x - p)(x - q)\), and standard form, \(y = ax^2 + bx + c\), are two ways of writing quadratic functions. In this target, the focus will be on comparing quadratic functions to vertex form.
In the Solving Quadratic Equations Unit, the connection between the graph of a quadratic function and its equation was introduced (click here to review). This concept focused on writing quadratic functions in both factored form and standard form based on the graph of a function. Factored form, \(y = a(x - p)(x - q)\), and standard form, \(y = ax^2 + bx + c\), are two ways of writing quadratic functions. In this target, the focus will be on comparing quadratic functions to vertex form.
Quadratic function written in vertex form: \(f(x) = a(x - h)^2 + k\)
The point, \((h, k)\), in the function represents the vertex of the graph of the function.
The point, \((h, k)\), in the function represents the vertex of the graph of the function.
Remember that the graph of a quadratic function is called a parabola. The vertex of the parabola is a point on the parabola where the direction changes. It is either the highest point (maximum) on the parabola, or the lowest point (minimum) on the parabola. The vertex of the parabola can also provide us with the range of the function. The domain of a quadratic function is always the set of all real numbers because the graph of the function will continue to the left and to the right spanning all \(x\)-values. The range of the function can be found by looking at the \(y\)-coordinate of the vertex. The range will be written as an inequality. If the parabola opens up, the range is the set of all possible numbers that are greater than or equal to the \(y\)-coordinate of the vertex. If the parabola opens down, then the range is the set of all possible numbers that are less than or equal to the \(y\)-coordinate of the vertex. Take a look at the graphs below.
Graph A is a parabola that opens up, therefore the vertex will be the lowest point on the graph, or the minimum. The \(y\)-coordinate of the vertex is \(-4\), so the range would be \({y > -4}\). Graph B is a parabola that opens down, which means that the vertex will be the highest point on the graph, or the maximum. Since the y-coordinate of the vertex is \(5\), and the graph opens down, the range would be \({y < 5}\).
Comparing Quadratic Functions
In order to compare the graphs of quadratic functions, we need a reference function to always refer or compare back to. The reference function for quadratic functions is \(f(x) = x^2\). We can rewrite this function in vertex form: \(f(x) = (x - 0)^2 + 0.\) The vertex will be at the origin because the values of \((h, k)\) are \((0, 0)\).
In the interactive graph below, the black graph represents the graph of the reference function \(f(x) = x^2\). Use the sliders for \(a, h,\) and \(k\), to explore how changing a quadratic function in vertex form affects the graph of the function as compared to \(f(x) = x^2\). Use the questions below to help guide your exploration.
Comparing Quadratic Functions
In order to compare the graphs of quadratic functions, we need a reference function to always refer or compare back to. The reference function for quadratic functions is \(f(x) = x^2\). We can rewrite this function in vertex form: \(f(x) = (x - 0)^2 + 0.\) The vertex will be at the origin because the values of \((h, k)\) are \((0, 0)\).
In the interactive graph below, the black graph represents the graph of the reference function \(f(x) = x^2\). Use the sliders for \(a, h,\) and \(k\), to explore how changing a quadratic function in vertex form affects the graph of the function as compared to \(f(x) = x^2\). Use the questions below to help guide your exploration.
- Use the slider to change the value of \(a\). What happens to the graph of the function as "\(a\)" becomes larger? Smaller? What happens when "\(a\)" is a negative?
- Use the slider to change the value of \(h\). How does the graph change as "\(h\)" changes?
- Use the slider to change the value of \(k\). How does the graph change as "\(k\)" changes?
Does this seem familiar? In the Library of Functions Unit, the concept of comparing absolute value functions was explored, and the roles of \(a, h,\) and \(k\), were introduced. These values play the same roles here when comparing quadratic functions. When comparing quadratic functions, there are these three components to consider. When you are describing the comparison between a quadratic function and the reference function \(f(x) = x^2\), you always want to keep in mind the values of the \(a, h,\) and \(k\), and the roles they play in the transformations of the graph. Use the following descriptions when comparing quadratic functions.
\(a\) value:
- if \(|a| > 1\), vertically stretched
- if \(0 < |a| < 1\), vertically compressed
- if \(a\) is positive: opens up
- if \(a\) is negative: opens down
- if adding an \(h\) value: vertex translates to the left
- if subtracting an \( h\) value: vertex translates to the right
- if adding a \(k\) value: vertex translates up
- if subtracting a \(k\) value: vertex translates down
Compare the following functions to the reference function \(f(x) = x^2\) by typing the equations into Demos.
Example 1: \(h(x) =\frac{1}{3}(x-2)^{2}+ 4\)
Solution:
For \(h(x)\)
Example 1: \(h(x) =\frac{1}{3}(x-2)^{2}+ 4\)
Solution:
For \(h(x)\)
- \(a\): positive \(\Large\frac{1}{3}\), so the graph would OPEN UP and be vertically COMPRESSED since the absolute value of \(\Large\frac{1}{3}\), \(\left|\Large\frac{1}{3}\right| = \Large\frac{1}{3}\), is between \(0\) and \(1\)
- \(h\): subtracting a \(2\) so the graph would be TRANSLATED to the RIGHT \(2\) units
- \(k\): adding a \(4\) so the graph would be TRANSLATED UP \(4\) units
- the vertex of \(h(x)\) is located at \((2, 4)\)
Example 2: \(j(x)=-4x^{2}+1\)
Solution:
For \(j(x)\)
Solution:
For \(j(x)\)
- \(a\): negative \(4\), so the graph would OPEN DOWN and be vertically STRETCHED since the absolute value of \(-4\), \(|-4| = 4\), is greater than \(1\)
- \(h\): there is no "\(h\)" value inside of parentheses so the graph would not be translated to the left or right
- \(k\): adding a \(1\) so the graph would be TRANSLATED UP \(1\) unit
- the vertex of \(j(x)\) is located at \((0, 1)\)
Example 3: \(k(x)=\frac{5}{4}(x+6)^{2}-3\)
Solution:
For \(k(x)\)
Solution:
For \(k(x)\)
- \(a\): positive \(\Large\frac{5}{4}\), so the graph would OPEN UP, and be vertically STRETCHED since the absolute value of \(\Large{5}{4}\), \(\left|\Large\frac{5}{4}\right| = \frac{5}{4}\), is greater than \(1\)
- \(h\): adding a \(6\) so the graph would be TRANSLATED to the LEFT \(6\) units
- \(k\): subtracting a \(3\) so the graph would be TRANSLATED DOWN \(3\) units
- the vertex of \(k(x)\) is located at \((-6, -3)\)
Quick Check
Compare the following functions to \(f(x) = x^2\). Create a sketch of each function.
1) \(j(x) = -2(x+4)^{2}-5\)
2) \(k(x)=\large\frac{1}{2}\normalsize(x-3)^{2}\)
Quick Check Solutions
Compare the following functions to \(f(x) = x^2\). Create a sketch of each function.
1) \(j(x) = -2(x+4)^{2}-5\)
2) \(k(x)=\large\frac{1}{2}\normalsize(x-3)^{2}\)
Quick Check Solutions