Vertex Form is a way of expressing a quadratic function in the form:
\(y = a(x-h)^2 + k\)
- \(x\) represents the dependent variable
- \(y\) represents the independent variable
- \(a\) represents the direction the parabola opens and its compression/stretch factor
- \(h\) represents the factor the parabola gets shifted horizontally
- \(k\) represents the factor the parabola gets shifted vertically
Vertex Form is most commonly used for finding the vertex of a given quadratic function.
For example, we can identify the vertex of the function \(y = (x - 2)^2 + 5\) as \((2, 5)\) since \(h = 2\) and \(k = 5\).
This can be graphically represented as such:
Identify the Vertex Form of a quadratic that has an Axis of Symmetry of \(-9\), and an Optimal Value of \(6\).
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We can identify the Axis of Symmetry (\(-9\)) and Optimal Value (\(6\)) as the x and y-coordinates of the vertex resepctively:
\(y = (x + 9)^2 + 6\)
Therefore, we can determine the equation in Vertex Form as \(y = (x + 9)^2 + 6\).
Identify the Vertex Form of a quadratic that passes through the points (1,5) and (3,29).
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In order to determine the equation, we need to identify the \(a\) and \(k\) values.
First, we can plug the both pairs of values into the formula \(y = ax^2 + k\) to get separate equations:
\(\text{Equation}_1: 5 = a(1)^2 + k\)
\(5 = a + k\)
\(\text{Equation}_2: 29 = a(3)^2 + k\)
\(29 = 9a + k\)
Next, we can use the elimination method to cancel out common terms. In this instance, we will be subtracting Equation 1 from Equation 2:
\(29 \textcolor{red}{- 5} = 9a + \cancel{k} \textcolor{red}{- (a + \cancel{k})}\)
\(24 = 8a\)
We can divide the remaining terms to solve for \(a\):
\(\cfrac{8a}{\textcolor{red}{8}} = \cfrac{24}{\textcolor{red}{8}}\)
\(a = 3\)
We can plug the value of \(a\) into one of the original equations to determine \(k\) and verify that \(a\) is correct. In this instance, we will be using \(5 = a(1)^2 + k\):
\(5 = 3(1)^2 + k\)
\(5 = 3 + k\)
\(k = 5 - 3\)
\(k = 2\)
Therefore, we can determine that \(a = 3\) and \(k = 2\).
Converting Vertex Form to Standard Form
- Rearrange the function if necessary so that all terms are on one side
- Simplify the expression by squaring the binomial
- Expand the expression using distributive property
- Further simplify by collecting like terms
Example
Convert \(y = 4(x - 2)^2 + 6\) to Standard Form.
Since all terms are already on one side, we can skip Step 1.
First, we can simplify the expression by squaring the binomial:
\(y = 4(x - 2)(x - 2) + 6\)
\(y = 4((x)(x) + (x)(-2) + (-2)(x) + (-2)(-2)) + 6\)
\(y = 4(x^2 - 2x - 2x + 4) + 6\)
\(y = 4(x^2 - 4x + 4) + 6\)
Next, we can expand the expression using distributive property:
\(y = (4)(x^2) - (4)(4x) + (4)(4) + 6\)
\(y = 4x^2 - 16x + 16 + 6\)
Finally, further simplify by collecting like terms:
\(y = 4x^2 - 16x + 22\)
Therefore, we can determine that \(y = 4(x - 2)^2 + 6\) in Standard Form is \(y = 4x^2 - 16x + 22\).
Convert \(y - 4 = \cfrac{1}{3}(x + 1)^2\) to Standard Form
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First, we can rearrange the equation by shifting all terms onto one side:
\(y = \cfrac{1}{3}(x + 1)^2 + 4\)
Next, we can simplify the expression by squaring the binomial:
\(y = \cfrac{1}{3}(x + 1)(x + 1) + 4\)
\(y = \cfrac{1}{3}((x)(x) + (1)(x) + (1)(x) + (1)(1)) + 4\)
\(y = \cfrac{1}{3}(x^2 + x + x + 1) + 4\)
\(y = \cfrac{1}{3}(x^2 + 2x + 1) + 4\)
Next, we can expand the expression using distributive property:
\(y = (\cfrac{1}{3})(x^2) - (\cfrac{1}{3})(2x) + (\cfrac{1}{3})(1) + 4\)
\(y = \cfrac{1}{3}x^2 - \cfrac{2}{3}x + \cfrac{1}{3} + 4\)
Finally, further simplify by collecting like terms:
\(y = \cfrac{1}{3}x^2 - \cfrac{2}{3}x + \cfrac{1}{3} + \cfrac{12}{3}\)
\(y = \cfrac{1}{3}x^2 - \cfrac{2}{3}x + \cfrac{13}{3}\)
Therefore, we can determine that \(y - 4 = \cfrac{1}{3}(x + 1)^2\) converted to Standard Form is \(y = \cfrac{1}{3}x^2 - \cfrac{2}{3}x + \cfrac{13}{3}\).
Converting Vertex Form to Factored Form
- Convert to Standard Form using the steps listed above
- Factor using various techniques learnt
A rocket travels according to the equation \(h = -4.9(t - 6)^2 + 182\) where \(h\) is the height, in metres, above the ground and \(t\) is the time, in seconds.
1. Sketch a graph of the rocket's motion
2. Find the maximum height of the rocket
3. How long does it take the rocket to reach its maximum height?
4. How high was the rocket above the ground when it was fired?
Show Answer
To assist in sketching the graph, we can first make a table of values to determine the main points:
| X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| Y |
5.6 |
59.5 |
103.6 |
137.9 |
162.4 |
177.1 |
182 |
177.1 |
162.4 |
137.9 |
103.6 |
59.5 |
5.6 |
We can now sketch the graph as such:
From looking at the graph, we can determine the parabola's vertex, which will help us identify the rocket's maximum height and the time it takes to reach that height.
In this instance, its maximum height is \(182\;[m]\) and it takes \(6\;[s]\) to reach it.
In order to prove this algebraically, we need to find the vertex of the parabola. To do this, we need to set \(x = 6\):
\(h = -4.9(6 - 6)^2 + 182\)
\(h = -4.9(0)^2 + 182\)
\(h = -4.9(0) + 182\)
\(h = 182\;[m]\)
Therefore, we can determine that the maximum height of the rocket is \(182\;[m]\).
In order to determine the height of the rocket when first launched, we need to set \(x = 0\):
\(h = -4.9(x - 6)^2 + 182\)
\(h = -4.9(-6)^2 + 182\)
\(h = -4.9(36) + 182\)
\(h = -176.4 + 182\)
\(h = 5.6\;[m]\)
Therefore, we can determine that the rocket's launch height was \(5.6\;[m]\).