Standard Form
Standard Form is a way of expressing a quadratic function in the form:
- \(x\) represents the independent variable
- \(y\) represents the dependent variable
- \(a\) represents the stretch factor of the function and determines whether it opens upward or downward
- \(c\) represents the y-intercept
where \(a ≠ 0\) and all coefficients are real numbers.
First, we can create a table of values to identify where each of the main points are located:
| x Values | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|
| y Values | 8 | 3 | 0 | -1 | 0 | 3 | 8 |
Next, we can create our graph based on these values:
Table of Values
Every quadratic equation is a relationship of \(x\) and \(y\) values. To create a table of values, we just have to pick a set of \(x\) values, substitute them into the equation and evaluate to get the \(y\)-values. Alternatively, you could read the points from the corrsponding graph.
Enter the \(a\), \(b\), and \(c\) values below to create a table of values:
Converting Standard Form to Factored Form
Converting Standard Form to Factored Form is called Factoring of Quadratics.
- Check if there are any common factors
- Determine \(2\) integers that result in the sum of \(b\) and the product of \(ac\)
- Factor the first and last pairs of terms separatately
First, we can identify \(2\) as a common factor across all terms. We can factor out the \(2\) to simplify the function:
Next, we need to determine \(2\) integers that result in the sum of \(b\) (6) and product of \(ac\) (5). We can create a table of values to make this process easier for ourselves:
| \(a\) Value | 1 | -1 | 2 | -2 |
|---|---|---|---|---|
| \(c\) Value | 5 | -5 | 3 | -3 |
| Sum | 6 | -6 | 5 | -5 |
| Product | 6 | 6 | 6 | 6 |
We can rewrite the function with these numbers in order to factor the 2 pairs of terms:
\(y = 2(x^2 + x + 5x + 5)\)
\(y = 2(x(x + 1) + 5(x + 1))\)
\(y = 2((x + 1) + (x + 5))\)
Therefore, \(2(x^2 + 6x + 5)\) converted to Factored Form is \(\boldsymbol{2((x + 1) + (x + 5))}\).
Converting Standard Form to Vertex Form
The process of converting a Quadratic Function in Standard Form to Vertex Form is identical to Completing the Square.
- Factor the \(a\) value (if any) out of the first \(2\) terms
- Find the special value \(\left(\cfrac{b}{2}\right)^2\)
- Add, the special number to the bracket
- Subtract the expression by the special number by placing it outside the brackets and multiplying it by \(a\)
- Simplify the constants outside the brackets
- Factor the trinomial inside the brackets as the square of a binomial
As \(a = 3\), we can factor it out of the first 2 terms:
Next, as \(b = 3\), we can use this value to find the special number:
Then, we can add the special number to the brackets and subtract the expression by the special number multiplied by factor \(a\). We can then simplify the constants:
\(y = 3(x^2 + 8x + 16) + 6 - (3)(16)\)
\(y = 3(x^2 + 8x + 16) + 6 - 48\)
\(y = 3(x^2 + 8x + 16) - 42\)
Finally, we can factor the trinomial inside the brackets by finding \(2\) numbers that result in the sum of the second term and the product of the third term:
\(y = 3(x² + 4x + 4x + 16) - 42\)
\(y = 3(x(x + 4) + 4(x + 4)) - 42\)
\(y = 3(x + 4)² - 42\)
Therefore, we can determine that function \(y = 3x^2 + 24x + 6\) converted to vertex form is \(\boldsymbol{y = 3(x + 4)² - 42}\).
A ball is thrown upward with an initial velocity of \(10\;[\text{m/s}]\). Its approximate height \(h\), in metres, above the ground after \(t\) seconds is given by the relation \(h = -5t^2 + 10t + 35\).
- Sketch a graph of the quadratic relation
- Find the maximum height of the ball
- Find how long it takes the ball to reach the maximum height
- Find when the ball is at ground level
i. First, we can create a table of values to better determine where each of the main points is located. As the time starts at \(0\), we won't need to account \(x < 0\):
| x Values | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y Values | 35 | 38.75 | 40 | 38.75 | 35 | 28.75 | 20 | 8.75 | -5 |
We can now create our graph using our table of values:
ii. In order to determine the maximum height of the ball, we need the vertex. We can identify the vertex by Completing the Square:
\(h = -5(t^2 - 2t) + 35\)
\(\left(\cfrac{b}{2}\right)^2 = \left(\cfrac{-2}{2}\right)^2 = (-1)^2 = 1\)
\(h = -5(t^2 - 2t + 1 - 1) + 35\)
\(h = -5(t^2 - 2t + 1) -5(-1) + 35\)
\(h = -5(t^2 - t - t + 1) + 5 + 35\)
\(h = -5[t(t-1)-1(t - 1)] + 40\)
\(h = -5[(t-1)^2] + 40\)
Therefore, we can determine that the maximum height of the ball is \(\textbf{40 [m]}\).
iii. By determining the vertex of the quadratic in the previous part, we can determine that it took \(\textbf{1 [s]}\) for the ball to reach its maximum height.
iv. We can determine when the ball is at ground level by first setting the original equation equal to \(0\):
We can then use the quadratic formula to determine the roots of the equation. These roots represent the potential times the ball reaches the ground level.
Since \(\textcolor{red}{a = -5}\), \(\textcolor{green}{b = 10}\), and \(\textcolor{blue}{c = 35}\), we can plug these values into the quadratic formula to solve for the roots:
\(x = \cfrac{-\textcolor{green}{b} \pm \sqrt{\textcolor{green}{b}^2 - 4\textcolor{red}{a}\textcolor{blue}{c}}}{2\textcolor{red}{a}}\)
\(x = \cfrac{-\textcolor{green}{10} \pm \sqrt{\textcolor{green}{10}^2 - 4(\textcolor{red}{-5})(\textcolor{blue}{35})}}{2(\textcolor{red}{-5})}\)
\(x = \cfrac{-10 \pm \sqrt{100 + 700}}{-10}\)
\(x = \cfrac{-10 \pm 20\sqrt{2}}{-10}\)
Using \(\pm\), we can determine the \(2\) roots either by adding or subtracting the values in the numerator.
We can determine the first root by adding the values in the numerator:
\(x_1 = \cfrac{-10 + 20\sqrt{2}}{-10}\)
\(x_1 = \cfrac{18.28427125}{-10}\)
\(x_1 \approx -1.83\;[\text{s}]\)
We can determine the second root by subtracting the values in the numerator:
\(x_2 = \cfrac{-10 - 20\sqrt{2}}{-10}\)
\(x_2 = \cfrac{-38.28427125}{-10}\)
\(x_2 = 3.83\;[\text{s}]\)
We can determine that the \(2\) potential times the ball is at ground level are \(\approx -1.83\;[\text{s}]\) and \(3.83\;[\text{s}]\).
Since the time can't be negative, we can determine that the ball is at ground level at \(\textbf{3.83 [s]}\).