Certainly! Let's convert the quadratic function [tex]\( f(x) = x^2 + 10x - 28 \)[/tex] to its vertex form step-by-step using the method of completing the square.
### Step-by-Step Conversion to Vertex Form
1. Start with the given quadratic function:
[tex]\[ f(x) = x^2 + 10x - 28 \][/tex]
2. Isolate the quadratic and linear terms:
[tex]\[ f(x) = x^2 + 10x - 28 \][/tex]
3. To complete the square, take the coefficient of [tex]\( x \)[/tex], which is 10, divide it by 2, and square the result:
[tex]\[ \left(\frac{10}{2}\right)^2 = 5^2 = 25 \][/tex]
4. Add and subtract this square inside the function. We'll show this step-by-step:
[tex]\[ f(x) = x^2 + 10x + 25 - 25 - 28 \][/tex]
Notice that adding and subtracting 25 keeps the equation balanced because [tex]\( +25 - 25 \)[/tex] equals 0.
5. Group the perfect square trinomial and the constants:
[tex]\[ f(x) = (x^2 + 10x + 25) - 53 \][/tex]
Here, [tex]\( -28 - 25 = -53 \)[/tex].
6. Factor the perfect square trinomial:
[tex]\[ (x^2 + 10x + 25) = (x + 5)^2 \][/tex]
Thus, we have:
[tex]\[ f(x) = (x + 5)^2 - 53 \][/tex]
7. Vertex form of the function:
[tex]\[ f(x) = (x + 5)^2 - 53 \][/tex]
### Verification
To verify that our vertex form is equivalent to the original quadratic form, we can expand the vertex form back into standard form and compare.
[tex]\[ (x + 5)^2 - 53 \][/tex]
[tex]\[ = (x + 5)(x + 5) - 53 \][/tex]
[tex]\[ = x^2 + 10x + 25 - 53 \][/tex]
[tex]\[ = x^2 + 10x - 28 \][/tex]
This matches our original function, [tex]\( f(x) = x^2 + 10x - 28 \)[/tex].
### Graph Comparison
To further verify, let's compare the graphs of the two functions:
1. Original Function:
[tex]\[ f(x) = x^2 + 10x - 28 \][/tex]
2. Vertex Form:
[tex]\[ f(x) = (x + 5)^2 - 53 \][/tex]
Both functions should have the same graph since they are equivalent algebraically.
### Conclusion
The vertex form of the quadratic function [tex]\( f(x) = x^2 + 10x - 28 \)[/tex] is:
[tex]\[ f(x) = (x + 5)^2 - 53 \][/tex]
By completing the square, identifying and factoring the perfect square trinomial, and verifying through algebraic expansion and graph comparison, we established that the two forms of the function are indeed equivalent.
