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

#1**+4 **

We want to get the equation into vertex form.

y = 34x^{2} - 6x + 15 Subtract 15 from both sides.

y - 15 = 34x^{2} - 6x Divide through by 34.

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

(y - 15)/34 + 9/1156 = x^{2} - 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**+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