The values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
<h3>How to rewrite in vertex form?</h3>
The equation is given as:
f(x) = x^2 + 12x + 6
Rewrite as:
x^2 + 12x + 6 = 0
Subtract 6 from both sides
x^2 + 12x = -6
Take the coefficient of x
k = 12
Divide by 2
k/2 = 6
Square both sides
(k/2)^2 = 36
Add 36 to both sides of x^2 + 12x = -6
x^2 + 12x + 36= -6 + 36
Evaluate the sum
x^2 + 12x + 36= 30
Express as perfect square
(x + 6)^2 = 30
Subtract 30 from both sides
(x + 6)^2 -30 = 0
So, the equation f(x) = x^2 + 12x + 6 becomes
f(x) = (x + 6)^2 -30
A quadratic equation in vertex form is represented as:
f(x) = a(x - h)^2 + k
Where:
Vertex = (h,k)
By comparison, we have:
(h,k) = (-6,-30)
Hence, the values of h and k when f(x) = x^2 + 12x + 6 is in vertex form is -6 and -30
Read more about quadratic functions at:
brainly.com/question/1214333
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So to solve this first you need to convert all the answers to decimals.
Site B 17/9 = 1.89 when you round up
Site D 4/3 = .75
Then look at all the numbers least to greatest.
D .75 m C 1.36 m A 1.60 m B 1.89 m
Answer D, C, A, B
Answer:
a = - 4, b = 5
Step-by-step explanation:
Expand the left side, then compare the coefficients of like terms.
- 3(2x² + ax + b)
= - 6x² - 3ax - 3b, compare to - 6x² + 12x - 15
Compare coefficients of x- terms
- 3a = 12 ( divide both sides by - 3 )
a = - 4
Compare constant terms
- 3b = - 15 ( divide both sides by - 3 )
b = 5
I have this same problem I need help too!!!
• Im sorry but we doesnt know the x number :( but dont worry to plot the diagram:
- substitute the number x (given) into the equation f(x)
- example: f(x) = x + 9 / x - 3
- then x given is 4
- therefore, f(x) = (4) + 9 / (4) - 3
- f(x) = 13 / 1 or 13
- then you plot at 13 on the graph :)
hope this helps ;/
