How do you find the vertex, axis of symmetry and intercepts of $f (x) = 2 x^{2} + x = 3$ $f(x)=2x^2+x=3$ ?

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How do you find the vertex, axis of symmetry and intercepts of $f (x) = 2 x^{2} + x = 3$ $f(x)=2x^2+x=3$ ?

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Answer 1 · 2017-12-21T11:40:05

$see explanation$ $"see explanation"$

Explanation:

$given a parabola in standard form$ $"given a parabola in "color(blue)"standard form"$

$∙ a x^{2} + b x + c x; a \neq 0$ $•color(white)ax^2+bx+c color(white)(x);a!=0$

$then the x-coordinate of the vertex which is also the$ $"then the x-coordinate of the vertex which is also the"$
$axis of symmetry is$ $"axis of symmetry is"$

$∙ x x_{vertex} = - \frac{b}{2 a}$ $•color(white)(x)x_(color(red)"vertex")=-b/(2a)$

$f (x) = 2 x^{2} + x - 3 is in standard form$ $f(x)=2x^2+x-3" is in standard form"$

$with a = 2, b = 1, c = - 3$ $"with "a=2, b=1, c=-3$

$\Rightarrow x_{vertex} = - \frac{1}{4}$ $rArrx_(color(red)"vertex")=-1/4$

$\Rightarrow axis of symmetry is x = - \frac{1}{4}$ $rArr"axis of symmetry is "x=-1/4$

$substitute this value into the equation for y-coordinate$ $"substitute this value into the equation for y-coordinate"$

$y_{vertex} = 2 {(- \frac{1}{4})}^{2} + (- \frac{1}{4}) - 3 = - \frac{25}{8}$ $y_(color(red)"vertex")=2(-1/4)^2+(-1/4)-3=-25/8$

$\Rightarrow vertex = (- \frac{1}{4}, - \frac{25}{8})$ $rArrcolor(magenta)"vertex "=(-1/4,-25/8)$

$for intercepts$ $"for "color(blue)"intercepts"$

$∙ let x = 0, in the equation for y-intercept$ $• " let x = 0, in the equation for y-intercept"$

$∙ let y = 0, in the equation for x-intercepts$ $• " let y = 0, in the equation for x-intercepts"$

$x = 0 \to y = - 3 \leftarrow y-intercept$ $x=0toy=-3larrcolor(red)"y-intercept"$

$y = 0 \to 2 x^{2} + x - 3 = 0$ $y=0to2x^2+x-3=0$

$\Rightarrow (x - 1) (2 x + 3) = 0$ $rArr(x-1)(2x+3)=0$

$\Rightarrow x = 1 and x = - \frac{3}{2} \leftarrow x-intercepts$ $rArrx=1" and "x=-3/2larrcolor(red)"x-intercepts"$
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}

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How do you find the vertex, axis of symmetry and intercepts of $f (x) = 2 x^{2} + x = 3$ $f(x)=2x^2+x=3$ ?

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How do you find the vertex, axis of symmetry and intercepts of $f (x) = 2 x^{2} + x = 3$ $f(x)=2x^2+x=3$ ?