The answer is a = 1, b = 2, and c = -3. How just by look at the points? C is intuitive, but I don't get the other points.

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3 Answers
Nov 14, 2017

#if a>0 =>"smile" or uuu like => min#
#if a<0 =>"sad" or nnn like => max#

#x_min=(-b)/(2a)#
#y_min=y_((x_min))#

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

Explanation:

just to explain #x=(-b)/(2a)#:

if you want to find the #x_min# or #x_max# you do #y'=0#, right?

Now, because we are dealing with the form of
#y=ax^2+bx+c#
the differentiate is always in the form of
#y'=2ax+b#

now we say (in general):
#y'=0#
#=> 2ax+b=0#
#=> 2ax=-b#
#=> x=(-b)/(2a)#

So as we see, the x_max or x_min is always #x=(-b)/(2a)#

Nov 14, 2017

#a=1,b=2,c=-3#

Explanation:

#"one possible approach"#

#c=-3larrcolor(red)"y-intercept"#

#• " sum of roots "=-b/a#

#• " product of roots "=ca#

#"here the roots are "x=-3" and "x=1#

#"that is where the graph crosses the x-axis"#

#rArr-3xx1=carArrca=-3rArra=-3/(-3)=1#

#rArr-b/a=-3+1=-2rArrb=2#

#rArry=x^2+2x-3#
graph{x^2+2x-3 [-10, 10, -5, 5]}

Nov 14, 2017

Bit wordy but work you way through it. Full explanation given.

Explanation:

Given the standardised form #y=ax^2+bx+c#

The curve at the bottom has the special name (what doesn't in maths) of Vertex.

If there are x-intercepts (where the graph crosses the x-axis) then the Vertex value of #x# is #1/2# way between

Looking at the graph the x-intercepts are at #x=-3 and x=1#

So the #x# value of the vertex is the average

#x_("vertex") = (-3+1)/2=-1#

This is what relates #x_("vertex")# to the equation.

Write as #y=a(x^2+b/ax)+c" "......................Equation(1)#

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

#-1=(-1/2)xxb/a#

Divide both side by #(-1/2)#

#color(brown)(2=b/a)#

Substitute into #Equation(1)# giving

#y=a(x^2+2x)+c" "....................Equation(1_a)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets pick a known point.
I choose the left hand x-intercept #->(x,y)=(-3,0)#

Known that #c=-3#

Substitution into #Equation(1_a)#

#y=a[color(white)("dd")x^2color(white)("dd")+color(white)("d")2xcolor(white)(()^2)]+c#

#0=a[(-3)^2+2(-3)]-3#

Add 3 to both sides and simplify the brackets

#3=9a-6a#

#color(brown)(3=3a=>a=1)#

Thus #color(brown)(2=b/a->2=b/1=>b=2)#

#y=ax^2+bx+c#

#color(magenta)(y=x^2+2x-3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that:

#y=a(x^2+b/ax)+c" ".........Equation(1)#

is the beginnings of completing the square.