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y = 4x² + 16x ; Take the derivative: y' = 8x + 16. Set equal to zero, and x = -2.
Substitute into the original: y = 4(-2)² + 16*(-2) = -16, so the vertex is (-2,16).
Or, factor to y = 4x(x + 4).
The zeros of this are x = 0 and x = -4.
Since a parabola is symmetric, the vertex is halfway between these, or x = -2, then substitute as above and get (-2,16).
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Find the vertex of y equals x2 plus 10x plus 22?
-2-5
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What is the vertex of the parabola y equals x2 plus 8x plus 5?
The vertex has a minimum value of (-4, -11)
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The equation is linear and so has no vertex.
What is the vertex of the function y equals x squared plus 4?
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What is the vertex of y equals x2-6x plus 6?
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How do you find the vertex of 2x2 plus 4 plus 1?
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
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x = -3y = -14
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(-3, -5)
