# How do you find the vertex and the intercepts for  4x^2 - 3x - 2 = 0?

Mar 21, 2016

$\textcolor{b l u e}{\text{at } y = 0 \ldots . x = \frac{3 + \sqrt{23}}{4} \mathmr{and} x = \frac{3 - \sqrt{23}}{4}}$

$\textcolor{b l u e}{\text{vertex} \to \left(x , y\right) \to \left(\frac{3}{8} , - \frac{41}{16}\right)}$

color(blue)(y_("intercept")=-2

#### Explanation:

There is a really cool way to find the axis of symmetry and hence ${x}_{\text{vertex}}$. Some you can do in your head!

Write the given equation as:$\text{ } y = 4 \left({x}^{2} - \frac{3}{4} x\right) - 2$

Now consider the $- \frac{3}{4}$ from $- \frac{3}{4} x$

Apply:$\text{ } \left(- \frac{1}{2}\right) \times \left(- \frac{3}{4}\right) = + \frac{3}{8}$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \frac{3}{8}}$
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color(blue)("Determin "y_("vertex"))
Substitute $x = \frac{3}{8}$ in the original equation

${y}_{\text{vertex}} = 4 {\left(\frac{3}{8}\right)}^{2} - 3 \left(\frac{3}{8}\right) - 2$

${y}_{\text{vertex}} = \frac{9}{16} - \frac{9}{8} - 2 = - 2 \frac{9}{16} \to - \frac{41}{16}$

$\textcolor{b l u e}{{y}_{\text{vertex}} = - 2 \frac{9}{16} \to - \frac{41}{16}}$

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$\textcolor{b l u e}{\text{Determine "y_("intercept}}$

The graph will cross the y-axis at x=0 so by substitution.

$y = 4 {\left(0\right)}^{2} - 3 \left(0\right) - 2$

$y = 0 - 0 - 2$

color(blue)(y_("intercept")=-2
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color(blue)("Determine "x_("intercept"))

There are not whole number factors for this equation so we are forced to use the equation method.

$y = a {x}^{2} + b x + c$

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

and: a=4; b=-3; c=-2 " " giving

$x = \frac{+ 3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(4\right) \left(- 2\right)}}{2 \left(4\right)}$

$x = \frac{3 \pm \sqrt{23}}{4}$

23 is a prime number so you can not do any more with it. So if you whish have a precise solution there is no choice other than to state:

$\textcolor{b l u e}{\text{at } y = 0 \ldots . x = \frac{3 + \sqrt{23}}{4} \mathmr{and} x = \frac{3 - \sqrt{23}}{4}}$

You could work out the decimals if you wish but these will not be precise!

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