Question

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.

Answer 1

`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)`

Answer 2

`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]}

Answer 3

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.