Give it a shot!
You can rewrite any quadratic function in standard form \( y(x) = ax^{2} + bx + c \), into vertex form, \( f\left( x \right) = a{(x - h)}^{2} + k \).
Expressing quadratic functions in vertex form makes it easier to create the graph of the parabola. Follow these steps to rewrite quadratic functions into vertex form.
| Step 1 | Isolate the c term by grouping \( \left( ax^{2} + bx \right) + c \). |
| Step 2 | Reduce so that the value of \( a = 1 \). It is not always necessary to complete this step. |
| Step 3 | Find the value of \( \left( \frac{b}{2} \right)^{2} \). |
| Step 4 | Carefully add the value of \( \left( \frac{b}{2} \right)^{2} \) both inside and outside the parentheses. Remember to keep the equation balanced. |
| Step 5 | Simplify and write the quadratic function in vertex form. |
Practice writing quadratic functions from standard form to vertex form by completing the activity below. Rewrite the function on each tab by completing the square. Be sure to check your work.
Write \( f\left( x \right) = x^{2} + 8x - 10 \) in vertex form.
\( f\left( x \right) = \left( x + 4 \right)^{2} - 26 \)
If you need help arriving at this answer, click the solution button.
Isolate the c term by grouping \( \left( ax^{2} + bx \right) + c \). |
\( f\left( x \right) = x^{2} + 8x - 10 \) \( f\left( x \right) = \left( x^{2} + 8x \right) - 10 \) |
Reduce so that the value of \( a = 1 \). |
For this function, \( a = 1 \), so you can skip this step. |
Find the value of \( \left( \frac{b}{2} \right)^{2} \). |
\( \left( \frac{8}{2} \right)^{2} = 4^{2} = 16 \) |
Carefully add the value of \( \left( \frac{b}{2} \right)^{2} \) both inside and outside the parentheses. |
\( f\left( x \right) = \left( x^{2} + 8x + 16 \right) - 10 \) You added 16 inside the parentheses. To keep the equation balanced, you need to subtract 16 from the c term. \( f\left( x \right) = \left( x^{2} + 8x + 16 \right) - 10 - 16 \) |
Simplify and write the quadratic function in vertex form. |
\( f\left( x \right) = \left( x^{2} + 8x + 16 \right) - 26 \) \( f\left( x \right) = \left( x + 4 \right)^{2} - 26 \) |
Write \( g\left( x \right) = 3x^{2} + 12x + 7 \) in vertex form.
\( g\left( x \right) = 3\left( x + 2 \right)^{2} - 5 \)
If you need help arriving at this answer, click the solution button.
Isolate the c term by grouping \( \left( ax^{2} + bx \right) + c \). |
\( g\left( x \right) = 3x^{2} + 12x + 7 \) \( g\left( x \right) = \left( {3x}^{2} + 12x \right) + 7 \) |
Reduce so that the value of \( a = 1 \). |
\( g\left( x \right) = \left( {3x}^{2} + 12x \right) + 7 \) \( g\left( x \right) = 3\left( x^{2} + 4x \right) + 7 \) |
Find the value of \( \left( \frac{b}{2} \right)^{2} \). |
\( \left( \frac{4}{2} \right)^{2} = 2^{2} = 4 \) |
Carefully add the value of \( \left( \frac{b}{2} \right)^{2} \) both inside and outside the parentheses. |
\( g\left( x \right) = 3\left( x^{2} + 4x + 4 \right) + 7 \) It looks like you added 4 inside the parentheses, but you must account for the 3 that is sitting in front of the parentheses. Any value outside the parentheses multiplies everything inside the parentheses. You actually added 3(4) = 12. Balance the equation by subtracting 12 from the constant. \( g\left( x \right) = 3\left( x^{2} + 4x + 4 \right) + 7 - 12 \) |
Simplify and write the quadratic function in vertex form. |
\( g\left( x \right) = 3\left( x^{2} + 4x + 4 \right) - 5 \) \( g\left( x \right) = 3\left( x + 2 \right)^{2} - 5 \) |
Write \( h\left( x \right) = 2x^{2} + 12x + 5 \) in vertex form.
\( h\left( x \right) = 2\left( x + 3 \right)^{2} - 13 \)
If you need help arriving at this answer, click the solution button.
Isolate the c term by grouping \( \left( ax^{2} + bx \right) + c \). |
\( h\left( x \right) = 2x^{2} + 12x + 5 \) \( h\left( x \right) = \left( {2x}^{2} + 12x \right) + 5 \) |
Reduce so that the value of \( a = 1 \). |
\( h\left( x \right) = \left( {2x}^{2} + 12x \right) + 5 \) \( h\left( x \right) = 2\left( x^{2} + 6x \right) + 5 \) |
Find the value of \( \left( \frac{b}{2} \right)^{2} \). |
\( \left( \frac{6}{2} \right)^{2} = 3^{2} = 9 \) |
Carefully add the value of \( \left( \frac{b}{2} \right)^{2} \) both inside and outside the parentheses. |
\( h\left( x \right) = 2\left( x^{2} + 6x + 9 \right) + 5 \) Remember to account for the 2 that is sitting in front of the parentheses. You actually added 2(9) = 18. Balance the equation by subtracting 18 from the constant. \( h\left( x \right) = 2\left( x^{2} + 6x + 9 \right) + 5 - 18 \) |
Simplify and write the quadratic function in vertex form. |
\( h\left( x \right) = 2\left( x^{2} + 6x + 9 \right) - 13 \) \( h\left( x \right) = 2\left( x + 3 \right)^{2} - 13 \) |
Question
When you rewrite a quadratic function from standard form to vertex form, the two functions are equivalent. Explain how to show that \( h\left( x \right) = 2x^{2} + 12x + 5 \) is the same as \( h\left( x \right) = 2\left( x + 3 \right)^{2} - 13 \).
The simplest way to show that the two functions are equivalent is to expand \( h\left( x \right) = 2\left( x + 3 \right)^{2} - 13 \). You can multiply the binomial \( (x + 3) \) by itself and simplify. Then multiply the resulting trinomial by 2. Finish by subtracting 13 from its like term, and you should arrive at \( h\left( x \right) = 2x^{2} + 12x + 5 \).
