How do you find the VERTEX of a parabola #y=x^2+9x+8#?

Answer 1

Your parabola is in the form #y=ax^2+bx+c# where you have: #a=1# #b=9# #c=8# The coordinates of the vertex are: #x_v=-b/(2a)=-9/2# #y_v=-(Delta)/(4a)=-(b^2-4ac)/(4a)=-49/4#

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Answer 2

Ethan Adkins

To find the vertex of the parabola ( y = x^2 + 9x + 8 ), you can use the formula ( x = \frac{-b}{2a} ), where ( a ) is the coefficient of ( x^2 ) (which is 1 in this case) and ( b ) is the coefficient of ( x ) (which is 9 in this case). So, ( x = \frac{-9}{2(1)} = -\frac{9}{2} ). Then, substitute ( x = -\frac{9}{2} ) into the equation to find the corresponding ( y ) value. Therefore, the vertex is ( (-\frac{9}{2}, \frac{-7}{4}) ).

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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.