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How do you find the VERTEX of a parabola #y= x^2 + 6x + 5#?

Answer 1

Genesis Allen

Complete the square to get the equation into vertex form:

#y = (x-(-3))^2+(-4)#

Then the vertex can be read as #(-3,-4)#

Any quadratic, such as #y = ax^2+bx+c#, can be completed squared to obtain:

#a(x+b/(2a))^2 + (c - b^2/(4a))#ax^2+bx+c

Since #a = 1#, #b = 6#, and #c = 5# in our example, we arrive at:

#x^2 + 6x + 5 = (x + 3)^2 + (5-3^2) = (x + 3)^2 + (5-9)#

#= (x+3)^2-4#

#y = (x+3)^2 - 4#, then.

The vertex #(h, k)# can be read off by replacing the #(x+3)# and the #-4# with #(x - (-3))#. Strictly speaking, the vertex form is #y = a(x-h)^2+k#.

(x-(-3))^2+(-4)#y =

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Answer 2

Jameson Allen

To find the vertex of the parabola y = x^2 + 6x + 5, you can use the formula for the x-coordinate of the vertex: x = -b/(2a), where a is the coefficient of x^2 and b is the coefficient of x in the equation. Then, plug the value of x into the equation to find the corresponding y-coordinate of the vertex.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.