What is the vertex of  y= -3x^2-x-2(3x+5)^2?

May 13, 2016

The vertex is at $\left(- \frac{61}{42} , - \frac{10059}{1764}\right)$ or $\left(- 1.45 , - 5.70\right)$

Explanation:

You can find the vertex from ANY of the three forms of a parabola: Standard, factored and vertex. Since it is simpler I'm going to converted this into standard form.

$y = - 3 {x}^{2} - x - 2 {\left(3 x + 5\right)}^{2}$

$y = - 3 {x}^{2} - x - 2 \cdot \left(9 {x}^{2} + 2 \cdot 5 \cdot 3 \cdot x + 25\right)$
$y = - 3 {x}^{2} - x - 18 {x}^{2} - 60 x - 50$
$y = - 21 {x}^{2} - 61 x - 50$

${x}_{v e r t e x} = \frac{- b}{2 a} = \frac{61}{2 \cdot \left(- 21\right)} = - \frac{61}{42} \cong - 1.45$

(you can prove this by either completing the square in general or averaging the roots found from the quadratic equation)

and then substituted it back into the expression to find ${y}_{v e r t e x}$

${y}_{v e r t e x} = - 21 \cdot {\left(- \frac{61}{42}\right)}^{2} - 61 \cdot \left(- \frac{61}{42}\right) - 50$

${y}_{v e r t e x} = \frac{- 21 \cdot 61 \cdot 61}{42 \cdot 42} + \frac{61 \cdot 61 \cdot 42}{42 \cdot 42} - \frac{50 \cdot 42 \cdot 42}{42 \cdot 42}$

${y}_{v e r t e x} = \frac{- 21 \cdot 61 \cdot 61 + 61 \cdot 61 \cdot 42 - 50 \cdot 42 \cdot 42}{42 \cdot 42}$

${y}_{v e r t e x} = - \frac{10059}{1764} \cong - 5.70$

The vertex is at $\left(- \frac{61}{42} , - \frac{10059}{1764}\right)$ or $\left(- 1.45 , - 5.70\right)$