Question

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

Answer 1

Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Step-by-step explanation:

Given

$f(x) = x^2 - x +2$

Required

The vertex

We have:

$f(x) = x^2 - x +2$

First, we express the equation as:

$f(x) = a(x - h)^2 +k$

Where

$Vertex = (h,k)$

So, we have:

$f(x) = x^2 - x +2$

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Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

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$f(x) = x^2 - x +2$

$f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2$

$f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}$

Expand

$f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}$

Factorize

$f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}$

Factor out x - 1/2

$f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}$

Compare to: $f(x) = a(x - h)^2 +k$

$h = \frac{1}{2}$

$k = \frac{7}{4}$

Hence:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Answer 2

Question: Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

Answer:  A or 1/2 , 7/4

Step-by-step explanation:

did it on an assignment on EDGE