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 B