Answer should be A.
GJ/MN = 7/4
hope it helps
The vertex form of the equation f(x) = x^2 - 3x, is f(x) = (x - 3/2)^2 - 9/5
<h3>How to rewrite the
quadratic function?</h3>
The quadratic function is given as:
f(x) = x^2 - 3x
Differentiate the function
f'(x) = 2x - 3
Set the function to 0
2x - 3 = 0
Add 3 to both sides
2x = 3
Divide by 2
x = 3/2
Set x = 3/2 in f(x) = x^2 - 3x
f(x) = 3/2^2 - 3 * 3/2
Evaluate
f(x) = -9/5
So, we have:
(x, f(x)) = (3/2, -9/5)
The above represents the vertex of the quadratic function.
This is properly written as:
(h, k) = (3/2, -9/5)
The vertex form of a quadratic function is
f(x) = a(x - h)^2 + k
So, we have:
f(x) = a(x - 3/2)^2 - 9/5
In f(x) = x^2 - 3x,
a = 1
So, we have:
f(x) = (x - 3/2)^2 - 9/5
Hence, the vertex form of the equation f(x) = x^2 - 3x, is f(x) = (x - 3/2)^2 - 9/5
Read more about vertex form at
brainly.com/question/24850937
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Answer:C
Step-by-step explanation:
I am not that sure.. but I think the answer is C.
