# How do you find the vertex and the intercepts for y= x^2-x-23 ?

May 26, 2017

x_("intercepts")color(white)(..)->x~~-4.32" and "x~~5.32 to 2 decimal places
$\text{ "x=1/2+-sqrt(93)/2" } \leftarrow$ exact values

${y}_{\text{intercept}} = - 23$

Vertex$\to \left(x , y\right) = \left(\frac{1}{2} , - \frac{93}{4}\right) \to \left(0.5 , - 23.25\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the x-intercepts}}$

Note that 23 is a prime number and as the coefficient of $x$ IS NOT $+ 1 - 23 = - 22$ the roots will be fractional. Thus use the formula

Given that: $y = a {x}^{2} + b x + c \text{ where } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case: $a = + 1 \text{; "b=-1"; } c = - 23$

$x = \frac{+ 1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(- 23\right)}}{2 \left(1\right)}$

$x = \frac{1 \pm \sqrt{93}}{2}$

93 is not prime so we can 'hunt' for squared values as factors of it. The whole number factors turn out to be 3 and 31. Both of which are prime so we are stuck with $\sqrt{93}$ as the exact value.

Thus $x = \frac{1}{2} \pm \frac{\sqrt{93}}{2} \text{ } \leftarrow$ exact values

$\text{ "x~~-4.321825..." and } x \approx 5.321825 \ldots$

$\text{ "x~~-4.32" and } x \approx 5.32$ to 2 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the vertex}}$

${x}_{\text{vertex}}$ will be midpoint between the x-intercepts

${x}_{\text{vertex}} \to \frac{- 4.32 + 5.32}{2} = 0.5$

As a check this is a 'sort of cheat method'

From y=ax^2+bx+c:" "x_("vertex")=(-1/2)xxb

$\text{ } \to \left(- \frac{1}{2}\right) \times \left(- 1\right) = + 0.5$

y_("vertex")=x^2-x-23" "=" "(0.5)^2-0.5-23
$\text{ } = - 23.25 = - \frac{93}{4}$

Vertex$\to \left(x , y\right) = \left(\frac{1}{2} , - \frac{93}{4}\right) \to \left(0.5 , - 23.25\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Set $x = 0$

$y = {x}^{2} - x - 23 \equiv > y = {0}^{2} + 0 - 23$

${y}_{\text{intercept}} = - 23$