# What is the vertex of y=4x^2+3x+18?

Apr 14, 2016

color(green)("Vertex"->(x,y)->(-3/8,279/16)

Notice the way I stick with fractions. Much more pricise than decimals.

#### Explanation:

There are various ways of doing this. I am going to show you one of them.

Write the equation as:

$y = 4 \left({x}^{2} + \frac{3}{4} x\right) + 18$

color(blue)("Determine "x_("vertex"))
Multiply the $\frac{3}{4}$ by $\left(- \frac{1}{2}\right)$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{3}{4} = - \frac{3}{8}}$

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Not that $- \frac{3}{8} = 0.375$
My graphing package has not rounded this properly to 2 decimal places
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color(blue)("Determine "y_("vertex"))

Substitute of $x$

${y}_{\text{vertex}} = 4 {\left(- \frac{3}{8}\right)}^{2} + 3 \left(- \frac{3}{8}\right) + 18$

${y}_{\text{vertex}} = 4 \left(+ \frac{9}{64}\right) - \frac{9}{8} + 18$

$\textcolor{b l u e}{{y}_{\text{vertex}} = \frac{9}{16} - \frac{9}{8} + 18 = \frac{279}{16}}$

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color(green)("Vertex"->(x,y)->(-3/8,279/16)