Write each quadratic function in the form f(x)=a(x−h)2+k, find the vertex
1. y=x^2-3x
2. y=x^2+x
3. y=2x^2−12x+22
4. y=−2x^2−4x−5
I will do one, you follow the same steps with the others.
3. y=2x^2−12x+22
step1: factor out the coefficient of the square term from the first two terms if it is other than +1,
I let the constant tag along at the end
y = 2(x^2 - 6x ) + 22
step2. take 1/2 of the coefficient of the x term inside the bracket, square it, then add it then subtract it within the bracket.
y = 2(x^2 - 6x + 9 - 9) + 22
step3. write the first 3 terms inside the bracket as a perfect square
y = 2( (x-3)^2 - 9) + 22
step4. distribute the outside factor over the two terms inside the bracket
y = 2(x-3)^2 - 18 + 22
step 5. simplify the constants at the end, and you are done!
y = 2(x-3)^2 + 4
so the vertex is (3,4)
To write each quadratic function in the form f(x) = a(x - h)^2 + k and find the vertex, we can use the process of completing the square. The vertex form of a quadratic equation allows us to easily determine the vertex coordinates (h, k).
Let's go through each function step by step:
1. y = x^2 - 3x
To rewrite in the vertex form, we need to complete the square. First, we take half of the coefficient of x (-3/2) and square it:
(-3/2)^2 = 9/4
Add this value inside the parentheses and subtract it outside the parentheses to keep the equation balanced:
y = (x^2 - 3x + 9/4) - 9/4
Simplify the perfect square trinomial inside the parentheses:
y = (x - 3/2)^2 - 9/4
So, the vertex form is f(x) = (x - 3/2)^2 - 9/4. The vertex is (3/2, -9/4).
2. y = x^2 + x
Following the same steps as above:
Take half of the coefficient of x (1/2) and square it:
(1/2)^2 = 1/4
Add this value inside the parentheses and subtract it outside the parentheses:
y = (x^2 + x + 1/4) - 1/4
Simplify the perfect square trinomial inside the parentheses:
y = (x + 1/2)^2 - 1/4
The vertex form is f(x) = (x + 1/2)^2 - 1/4. The vertex is (-1/2, -1/4).
3. y = 2x^2 - 12x + 22
Start by factoring out the common factor from the quadratic coefficient (2):
y = 2(x^2 - 6x) + 22
Take half of the coefficient of x (-6/2 = -3) and square it:
(-3)^2 = 9
Add this value inside the parentheses and subtract it outside:
y = 2(x^2 - 6x + 9 - 9) + 22
Simplify the perfect square trinomial inside the parentheses:
y = 2((x - 3)^2 - 9) + 22
Distribute 2 to both terms inside the parentheses:
y = 2(x - 3)^2 - 18 + 22
Combine like terms:
y = 2(x - 3)^2 + 4
The vertex form is f(x) = 2(x - 3)^2 + 4. The vertex is (3, 4).
4. y = -2x^2 - 4x - 5
Start by factoring out the common factor from the quadratic coefficient (-2):
y = -2(x^2 + 2x) - 5
Take half of the coefficient of x (2/2 = 1) and square it:
(1)^2 = 1
Add this value inside the parentheses and subtract it outside:
y = -2(x^2 + 2x + 1 - 1) - 5
Simplify the perfect square trinomial inside the parentheses:
y = -2((x + 1)^2 - 1) - 5
Distribute -2 to both terms inside the parentheses:
y = -2(x + 1)^2 + 2 - 5
Combine like terms:
y = -2(x + 1)^2 - 3
The vertex form is f(x) = -2(x + 1)^2 - 3. The vertex is (-1, -3).
By completing the square and rewriting the quadratic equations in vertex form, we can easily determine the vertex coordinates of each function.