The equation of a parabola is given. y=34x^2−6x+15
What are the coordinates of the vertex of the parabola?
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)
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