How do you evaluate 2x − 4y + 3x^2 + 7y − 16x?

1 Answer
May 2, 2016

see explanation
Simplified version ->color(blue)(+y=-x^2+1/3x)"

Explanation:

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

color(blue)("Simplifying the expression")

Collecting like terms:

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

-14x+3y+3x^2

Order by degree (power that x is taken to)

3x^2-14x+3y

Set as equal to zero

3x^2-14x+3y=0

Subtract 3y from both sides

-3y=3x^2-14x

Divide both sides by 3

-y=3/3x^2-14/3 x

Multiply by (-1)

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

At x=0" "->" "y=-(0)^2+1/3(0) => y= 0

Suppose we set x=1/3

Then y=x^2+1/3x" "->" "y=0=(-1/3xx1/3)+(1/3xx1/3)

color(blue)(=> x_("intercepts")-> x= 0" and "x=1/3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine y intercept")

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

=>color(blue)( y_("intercept")->y=0)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine vertex")

Consider the standard form " "y=ax^2+bx+c

Write this as" "y=a(x^2+b/ax)+c

Then x_("vertex")= (-1/2)xxb/a

In your case a=-1

so color(blue)(x_("vertex")=(-1/2)xxb -> (-1/2)xx-1/3 =+1/6)

Substitute x=1/6" in "y=-x^2+1/3x

color(blue)(=>y_("vertex")=-(1/6)^2+(1/3xx1/6)=1/18-1/36 = 1/36)

color(blue)("Vertex "->" "(x,y)=(1/6,1/36))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony BTony B