What is the vertex of $y = 4 x^{2} + 3 x + 18$ $y=4x^2+3x+18$ ?

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What is the vertex of $y = 4 x^{2} + 3 x + 18$ $y=4x^2+3x+18$ ?

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Answer 1 · 2016-04-14T16:46:03

$Vertex \to (x, y) \to (- \frac{3}{8}, \frac{279}{16})$ $color(green)("Vertex"->(x,y)->(-3/8,279/16)$

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:

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

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

$x_{vertex} = (- \frac{1}{2}) \times \frac{3}{4} = - \frac{3}{8}$ $color(blue)(x_("vertex")=(-1/2)xx3/4= -3/8)$

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

Substitute of $x$ $x$

$y_{vertex} = 4 {(- \frac{3}{8})}^{2} + 3 (- \frac{3}{8}) + 18$ $y_("vertex")=4(-3/8)^2+3(-3/8)+18$

$y_{vertex} = 4 (+ \frac{9}{64}) - \frac{9}{8} + 18$ $y_("vertex")=4(+9/64)-9/8+18$

$y_{vertex} = \frac{9}{16} - \frac{9}{8} + 18 = \frac{279}{16}$ $color(blue)(y_("vertex")=9/16-9/8+18 = 279/16)$

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$Vertex \to (x, y) \to (- \frac{3}{8}, \frac{279}{16})$ $color(green)("Vertex"->(x,y)->(-3/8,279/16)$

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What is the vertex of $y = 4 x^{2} + 3 x + 18$ $y=4x^2+3x+18$ ?

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