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Which point gives the vertex of f(x) = -x2 + 4x - 3?
A) (2,-1)
B) (2.1)
C) (2,-13)
D) (-2,-13)


Sagot :

Answer:

B

Step-by-step explanation:

There are several ways to find the vertex of function such as formula, completing the square or differential.

I will use the formula to find the vertex.

We are given the function:

[tex] \displaystyle \large{f(x) = - {x}^{2} + 4x - 3}[/tex]

Vertex Formula

Let (h,k) = vertex

[tex] \displaystyle \large{ \begin{cases} h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a} \end{cases}}[/tex]

From the function, compare the coefficients:

[tex] \displaystyle \large{a {x}^{2} + bx + c = - {x}^{2} + 4x - 3}[/tex]

  • a = -1
  • b = 4
  • c = -3

Therefore:-

[tex] \displaystyle \large{ \begin{cases} h = - \frac{4}{2( - 1)} \\ k = \frac{4( - 1)( - 3)- {4}^{2} }{4( - 1)} \end{cases}}[/tex]

Then evaluate for h-value and k-value.

[tex] \displaystyle \large{ \begin{cases} h = - \frac{4}{ - 2} \\ k = \frac{4( 3)- 16}{ - 4} \end{cases}} \\ \displaystyle \large{ \begin{cases} h = 2 \\ k = \frac{12 - 16}{ - 4} \end{cases}} \\ \displaystyle \large{ \begin{cases} h = 2 \\ k = \frac{ - 4}{ - 4} \end{cases}} \\ \displaystyle \large{ \begin{cases} h = 2\\ k = 1\end{cases}}[/tex]

Therefore the vertex is (h,k) = (2,1)

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