# How do you find the vertex and the intercepts for t(x) = 3(x² + 2x) + 5?

May 7, 2016

${y}_{\text{intercept}} \to \left(x , y\right) \to \left(0 , 5\right)$

${x}_{\text{intercept}} \in \mathbb{C} \to x = - 1 \pm \frac{\sqrt{6}}{3} \textcolor{w h i t e}{. .} i$

#### Explanation:

Given:$\text{ } t \left(x\right) = 3 \left({x}^{2} + 2 x\right) + 5$

This is part way towards converting standard equation form to vertex form.

$\textcolor{b l u e}{\text{Determine the y-intercept}}$

The constant of 5 is the same as the constant $c$ in $y = a {x}^{2} + b x + c$ the y-intercept is $y = c = 5$

$\textcolor{b l u e}{\implies {y}_{\text{intercept}} \to \left(x , y\right) \to \left(0 , 5\right)}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the x-intercepts for } x \in \mathbb{R}}$

First of all lets make sure they exist.

Consider the method of solving using the formula where

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The part that is $\sqrt{{b}^{2} - 4 a c}$ is called the determinant.

From $\sqrt{{b}^{2} - 4 a c} \text{ ; if } {b}^{2} - 4 a c < 0$ then there are no x-intercepts in the 'set' of numbers that are called 'Real'. However there are solutions that are in the 'set' of numbers called 'Complex'.

$\textcolor{b r o w n}{y = a {x}^{2} + b x + c} \textcolor{b l u e}{\text{ "->" } y = 3 {x}^{2} + 6 x + 5}$

color(brown)(=>sqrt(b^2-4ac)color(blue)(" "->" "sqrt(6^2-4(3)(5))

$\sqrt{36 - 60}$ and $36 - 60 < 0$

color(blue)("Thus "x_("intercept") !in RR)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Slightly higher level of mathematics

$\textcolor{b l u e}{\text{Determine the x-intercepts for " x in CC " (complex numbers)}}$

Completing the square gives:

Standard form$\text{ "y=ax^2+bx+c" "->" } y = 3 {x}^{2} + 6 x + 5$

$\text{ " ->" } a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k$

$t \left(x\right) = 3 \left({x}^{2} + 2 x\right) + 5 \text{ " ->" } t \left(x\right) = 3 {\left(x + 1\right)}^{2} + 5 + k$

where $k = \left(- 1\right) \times a {\left(\frac{b}{2 a}\right)}^{2}$

$k = \left(- 1\right) \times \left(3 \times {1}^{2}\right) = - 3$

$\implies t \left(x\right) = 3 {\left(x + 1\right)}^{2} + 2$

Set to zero
$\implies 0 = 3 {\left(x + 1\right)}^{2} + 2$

${\left(x + 1\right)}^{2} = - \frac{2}{3}$

$x = \pm \sqrt{- 1 \times \frac{2}{3}} \textcolor{w h i t e}{.} - 1$

$x = - 1 \pm \sqrt{\frac{2}{3}} \textcolor{w h i t e}{. .} \left(i\right)$

But $\sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$

color(blue)(x=-1+-sqrt(6)/3color(white)(..)i