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Find the vertex of this parabola:
y = -x2 + 6x – 15

Sagot :

subtomex0

[tex]y = - x {}^{2} + 6x - 15[/tex]

[tex]y {}^{ \gamma } = - 2x + 6[/tex]

[tex]y {}^{ \gamma } = 0[/tex]

[tex] - 2x + 6 = 0[/tex]

[tex] - 2x = - 6[/tex]

[tex]2x = 6[/tex]

[tex]x = 3[/tex]

Substitute x=3 in y=-x²+6x-15

[tex]y = - (3) {}^{2} + 6(3) - 15[/tex]

[tex]y = - 9 + 18 - 15[/tex]

[tex]y = - 9 + 3[/tex]

[tex]y = - 6[/tex]

Vertex: (3,-6)

hello!

[tex]==============================[/tex]

In order to find the vertex of a parabola, we use the following formula:

[tex]\mathfrak{\displaystyle\frac{-b}{2a} }[/tex]

Please remember that a parabola looks like this:

[tex]\tt{y=ax^2+bx+c}[/tex]

Now you know what a and b stand for.

In this case, a = -1, and b is 6:

[tex]\mathfrak{\displaystyle\frac{-6}{2(-1)} }[/tex]

Simplify:

[tex]\mathfrak{\displaystyle\frac{-6}{-2}}[/tex]

[tex]\mathfrak{3}\\=======================================[/tex]

Notes:

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Answered by

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