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The equation of a parabola is given. y=34x^2−6x+15

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The equation of a parabola is given. y=34x^2−6x+15

What are the coordinates of the vertex of the parabola?

Kakarot_15  May 2, 2017
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#1
+5552
+4

We want to get the equation into vertex form.

y = 34x2 - 6x + 15            Subtract 15 from both sides.

y - 15 = 34x2 - 6x             Divide through by 34.

(y - 15)/34 = x2 - 3x/17     Add ( [3/17]/2 )2 , or 9/1156 , to both sides.

(y - 15)/34 + 9/1156 = x2 - 3x/17 + 9/1156       Factor the right side.

(y - 15)/34 + 9/1156 = (x - 3/34)2                     Multiply through by 34.

y - 15 + 9/34 = 34(x - 3/34)2

y - 501/34 = 34(x - 3/34)2

So...when the equation is written in this form, we can see that the vertex is located at (3/34 , 501/34)

hectictar  May 2, 2017
#2
+79798
+2

Thanks, hectictar.....here's another method.....

y=34x^2−6x+15

In the form   y  = Ax^2 + Bx + C  the x coordinate of the vertex  is given by :

-B  / [ 2A ]   =    - (-6)  / [ 2 * 34 ]   =   6/ 68   =   3/34

Subbing this back into the function, the y coordinate of the vertex is :

34 (3/34)^2  - 6 ( 3/34)  + 15   =

9 / 34   -  18/34   +  15  =

-9/34  + 510/34  =

501 / 34

CPhill  May 2, 2017

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