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Find the vertex for the parabola whose equation is given by writing the equation in the form y = ax² +bx+c.y=(x-3)²-5***(Type an ordered pair.)The vertex is

Sagot :

Given: A parabola equation

[tex]y=(x-3)^2-5[/tex]

Required: To find the vertex of the given parabola.

Explanation: The given equation can be written as

[tex]\begin{gathered} y=x^2+9-6x-5 \\ y=x^2-6x+4 \end{gathered}[/tex]

Now the general equation is of the form

[tex]y=ax^2+bx+c[/tex]

Comparing both equations we get,

[tex]\begin{gathered} a=1 \\ b=-6 \\ c=4 \end{gathered}[/tex]

The x-coordinate of the vertex is

[tex]\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{-(-6)}{2(1)} \\ x=3 \end{gathered}[/tex]

For the y coordinate of the vertex, put x=-3 in the equation of parabola

[tex]\begin{gathered} y=(3)^2-6(3)+4 \\ y=-5 \end{gathered}[/tex]

Hence the coordinate of the vertex is (-3,31)

Final Answer: The vertex is (3,-5).

View image EmigdioF530842
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