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How do you write the Vertex form equation of the parabola #y=2x^2+10+9#?

Answer 1

Nathan Axel

#y=2(x+5/2)^2-7/2#

Assuming you meant #10color(red)(x)#, not #10#

#y=2x^2+10x+9#

#y=2[x^2+5x+9/2]#

#y=2[(x+5/2)^2-25/4+9/2]#

#y=2[(x+5/2)^2-7/4]#

#y=2(x+5/2)^2-7/2#

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Answer 2

Eva Adamson

To write the vertex form equation of the parabola ( y = 2x^2 + 10x + 9 ), complete the square. First, factor out the coefficient of ( x^2 ) from the quadratic term: ( y = 2(x^2 + 5x) + 9 ). Then, add and subtract ((\frac{5}{2})^2 = \frac{25}{4}) inside the parenthesis to complete the square: ( y = 2(x^2 + 5x + \frac{25}{4} - \frac{25}{4}) + 9 ). Simplify the expression inside the parenthesis: ( y = 2((x + \frac{5}{2})^2 - \frac{25}{4}) + 9 ). Distribute 2: ( y = 2(x + \frac{5}{2})^2 - \frac{25}{2} + 9 ). Combine constants: ( y = 2(x + \frac{5}{2})^2 - \frac{7}{2} ). Therefore, the vertex form equation of the parabola is ( y = 2(x + \frac{5}{2})^2 - \frac{7}{2} ).

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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.