What is the vertex of $y = 3 x^{2} - 4 x - 12$ $y= 3x^2 - 4x - 12$ ?

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Answer 1 · 2018-08-10T08:40:46

Vertex is $(\frac{2}{3}, \frac{96}{9})$ $(2/3,96/9)$

To find vertex convert the equation $y = 3 x^{2} - 4 x - 12$ $y=3x^2-4x-12$ to the form

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ and then vertex is $(h, k)$ $(h,k)$ .

We have $y = 3 x^{2} - 4 x - 12$ $y=3x^2-4x-12$

= $3 (x^{2} - 2 \times \frac{2}{3} x + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2}) - 12$ $3(x^2-2xx2/3x+(2/3)^2-(2/3)^2)-12$

= $3 (x^{2} - 2 \times \frac{2}{3} x + {(\frac{2}{3})}^{2}) - 3 \cdot \frac{4}{9} + 12$ $3(x^2-2xx2/3x+(2/3)^2)-3*4/9+12$

= $3 {(x - \frac{2}{3})}^{2} + \frac{96}{9}$ $3(x-2/3)^2+96/9$

Hence, vertex is $(\frac{2}{3}, \frac{96}{9})$ $(2/3,96/9)$

What is the vertex of y= 3x^2 - 4x - 12y=3x2−4x−12y= 3x^2 - 4x - 12?