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put the quadratic into vertex form and state the coordinates of the vertex y = x² - 6x + 45

Sagot :

[tex]\begin{gathered} y=(x-3)^2-(-3)^2+45=(x-3)^2-9+45=(x-3)^2+36 \\ \Rightarrow y-36=(x-3)^2 \end{gathered}[/tex]

Alternatively

Vertex coordinate =(h, k)

h= -b/2a

k = f(h)

From x^2 - 6x + 45

b= -6, a= 1

[tex]h\text{ = }\frac{-b}{2a}\text{ =}\frac{-(-6)_{}}{2(1)}\text{ =}\frac{6}{2}=\text{ 3}[/tex]

k = f(h)

Substitute h in x^2 -6x +45

[tex]\begin{gathered} k=f(3)=3^2\text{ - 6(3) + 45} \\ =\text{ 9-18 +45} \\ =\text{ 54-18=36} \end{gathered}[/tex]

The vertex coordinate is ( 3, 36)

The vertex form is

[tex]y=(x-3)^2\text{ + 36}[/tex]

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