How do you find the important points to graph y=-x^2-3x+2?

1 Answer
Aug 14, 2018

See explanation

Explanation:

Given: y=-x^2-3x+2
Compare to y=ax^2+bx+c

As the x^2 term in negative the general shape is nn

color(red)("The y-intercept "=c=+2 ->" point" (x,y)->(0,2))

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Write in the form y=a(x^2+b/ax)+c

In this case a=-1 and b=-3 giving:

y=-1(x^2+3x)+2

x_("vertex")=[color(white)(".")(-1/2)xx(b/a)color(white)(".")] -> [color(white)(".")(-1/2)xx(-3)/(-1)color(white)(".")] = -3/2

So by substitution:

y_("vertex")=-(-3/2)^2-3(-3/2)+2

y_("vertex")=-9/4+9/2+2 = 4 1/4->17/4

color(red)("Vertex"->(x,y)=(-3/2,17/4))
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Notice that (-1)xx(-2)=+2 larr c
and that (-1)+(-2) = -3larr b

so initially we would think that we have the factorisation. However the negative x^2 gives us a problem. Perhaps we can 'force' our initial thoughts to give us the correct form.

Set y=(x-1)(x-2)color(white)("dd") ->color(white)("dd") y=x^2-3x+2 larr" Fail"

Lets try:
y=(-x-1)(x-2)color(white)("dd")->color(white)("dd") y=-x^2+x+2larr" Fail"

Ok! Looks as though we do not have whole number factorisation. So lets use the formula x=(-b+-sqrt(b^2-4ac))/(2a)

y=color(white)("dd.d")ax^2+bx+c
y=(-1)x^2-3x+2

So a=-1; b=-3 and c=+2 giving:

x=(+3+-sqrt((-3)^2-4(-1)(+2)))/(2(-1))

x=-3/2+-sqrt(17)/2 larr" Exact solution"

As 17 is a prime number we can not simplify this any further.

Approximate solution:

x=-1.5+-2.062 to 3 decimal places giving:

color(red)(x_("intercept")~~ -3.562 and +0.562" to 3 decimal places")

Tony BTony B