How do you evaluate $2 x - 4 y + 3 x^{2} + 7 y - 16 x$ $2x − 4y + 3x^2 + 7y − 16x$ ?

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How do you evaluate $2 x - 4 y + 3 x^{2} + 7 y - 16 x$ $2x − 4y + 3x^2 + 7y − 16x$ ?

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Answer 1 · 2016-05-02T10:43:27

see explanation
Simplified version $\to + y = - x^{2} + \frac{1}{3} x$ $->color(blue)(+y=-x^2+1/3x)"$

Explanation:

$Depends on what is meant by 'evaluate'$ $color(magenta)("Depends on what is meant by 'evaluate'")$

$Simplifying the expression$ $color(blue)("Simplifying the expression")$

Collecting like terms:

$(2 x - 16 x) + (7 y - 4 y) + (+ 3 x^{2})$ $(2x-16x)+(7y-4y)+(+3x^2)$

$- 14 x + 3 y + 3 x^{2}$ $-14x+3y+3x^2$

Order by degree (power that $x$ $x$ is taken to)

$3 x^{2} - 14 x + 3 y$ $3x^2-14x+3y$

Set as equal to zero

$3 x^{2} - 14 x + 3 y = 0$ $3x^2-14x+3y=0$

Subtract $3 y$ $3y$ from both sides

$- 3 y = 3 x^{2} - 14 x$ $-3y=3x^2-14x$

Divide both sides by 3

$- y = \frac{3}{3} x^{2} - \frac{14}{3} x$ $-y=3/3x^2-14/3 x$

Multiply by (-1)

$+ y = - x^{2} + \frac{1}{3} x This may be as far as you need to go!$ $color(blue)(+y=-x^2+1/3x)" "color(magenta)("This may be as far as you need to go!")$
'~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the x-intercepts$ $color(blue)("Determine the x-intercepts")$

At $x = 0 \to y = - {(0)}^{2} + \frac{1}{3} (0) \Rightarrow y = 0$ $x=0" "->" "y=-(0)^2+1/3(0) => y= 0$

Suppose we set $x = \frac{1}{3}$ $x=1/3$

Then $y = x^{2} + \frac{1}{3} x \to y = 0 = (- \frac{1}{3} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{3})$ $y=x^2+1/3x" "->" "y=0=(-1/3xx1/3)+(1/3xx1/3)$

$\Rightarrow x_{intercepts} \to x = 0 and x = \frac{1}{3}$ $color(blue)(=> x_("intercepts")-> x= 0" and "x=1/3)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine y intercept$ $color(blue)("Determine y intercept")$

Given that one of the x intercepts is at $x = 0$ $x=0$
Then at $x = 0$ $x=0$ it is also the case that $y = 0$ $y=0$

$\Rightarrow y_{intercept} \to y = 0$ $=>color(blue)( y_("intercept")->y=0)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine vertex$ $color(blue)("Determine vertex")$

Consider the standard form $y = a x^{2} + b x + c$ $" "y=ax^2+bx+c$

Write this as $y = a (x^{2} + \frac{b}{a} x) + c$ $" "y=a(x^2+b/ax)+c$

Then $x_{vertex} = (- \frac{1}{2}) \times \frac{b}{a}$ $x_("vertex")= (-1/2)xxb/a$

In your case $a = - 1$ $a=-1$

so $x_{vertex} = (- \frac{1}{2}) \times b \to (- \frac{1}{2}) \times - \frac{1}{3} = + \frac{1}{6}$ $color(blue)(x_("vertex")=(-1/2)xxb -> (-1/2)xx-1/3 =+1/6)$

Substitute $x = \frac{1}{6} in y = - x^{2} + \frac{1}{3} x$ $x=1/6" in "y=-x^2+1/3x$

$\Rightarrow y_{vertex} = - {(\frac{1}{6})}^{2} + (\frac{1}{3} \times \frac{1}{6}) = \frac{1}{18} - \frac{1}{36} = \frac{1}{36}$ $color(blue)(=>y_("vertex")=-(1/6)^2+(1/3xx1/6)=1/18-1/36 = 1/36)$

$Vertex \to (x, y) = (\frac{1}{6}, \frac{1}{36})$ $color(blue)("Vertex "->" "(x,y)=(1/6,1/36))$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B Tony B

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How do you evaluate $2 x - 4 y + 3 x^{2} + 7 y - 16 x$ $2x − 4y + 3x^2 + 7y − 16x$ ?

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