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F (x)-x^2+6x-7 at x=2

User Gobernador
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Consider the vertex form of the parabola.y=a(x+h)2+ky=a(x+h)2+kRewrite the function in terms of xx and yy.y=x2+6x−7y=x2+6x-7Complete the square on the right side of the equation.(x+3)2−16(x+3)2-16Reorder the right side of the equation to match the vertex form of a parabola.y=(x+3)2−16y=(x+3)2-16Since 33 does not contain the variable to solve for, move it to the right side of the equation by subtracting 33 from both sides.x=−3x=-3Use the vertex form, y=a(x−h)2+ky=a(x-h)2+k, to determine the values of aa, hh, and kk.a=1a=1h=−3h=-3k=−16k=-16Since the value of aa is positive, the parabola opens up.Opens UpFind the vertex (h,k)(h,k).(−3,−16)(-3,-16)Find pp, the distance from the vertex to the focus.1414Find the focus.(−3,−634)(-3,-634)Find the axis of symmetry by finding the line that passes through the vertex and the focus.x=−3x=-3Find the directrix.y=−654y=-654Use the properties of the parabola to analyze and graph the parabola.Direction: Opens UpVertex: (−3,−16)(-3,-16)Focus: (−3,−634)(-3,-634)Axis of Symmetry: x=−3x=-3Directrix: y=−654

I sorta shortened the steps a bit if you need me to explain any more steps I will just ask.
I am sure you probably went over some of this in your class that is why I skipped a few steps so you would at least learn a little bit hehe.

I really hope this helps you.

User Jeremy Meadows
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