Question

Please help - and explain

Answer 1

interesting

I'm going to call the function for the graph of A, A(x) and the function for the graph of B, B(x),and the function for the graph of C, C(x)

we know that when A(x)=0 (and x>0), C(x)=B(x) at that value of x

also, when B(x)=0, A(x)=C(x) at that value of x

we need to find the intersection points

a quadratic function can be written in form y=a(x-b)(x-c) where the roots or zeroes are at x=b and x=c and a is a constant. since we can only find 2 points, we will assume a=1 since in the other quadratic function, a=1 as well

let's use our 1st discovery first

when A(x)=0, (and x>0), C(x)=B(x)

find zeroes of A(x)

$A(x)=x^2+2x-3$

find where it equals 0

$0=x^2+2x-3$, factor, what 2 numbers mulitly to get -3 and add to get 2

0=(x-1)(x+3)

set equal to 0

0=x-1, x=1

0=x+3, x=-3

we are looking for positive root, so at x=1, C(x)=B(x)

meaining C(1)=B(1), 3(1)+3=B(1), 3+3=B(1), B(1)=6, that's one point

2nd thing

when B(x)=0, A(x)=C(x)

solving A(x)=C(x) for x<0 (since it's to the left of y axis)

$x^2+2x-3=3x+3$

$x^2-x-6=0$

factor, what 2 numbers multiply to get -6 and add to get -1

(x-3)(x+2)=0

set equal to 0

x-3=0, x=3

x+2=0, x=-2

we want x<0 so at x=-2

find y value, C(3)=3(-2)+3=-6+3=-3

we know that B(x) has a zero at (-3,0)

we also know that B(x) passes through the point (1,6)

if a quadratic function has zeroes at m and n, then it can be factored into form y=a(x-m)(x-n) where a is a constant, we assume a=1 because otherwise we can't find the function since we only know 2 points

we know that one zero is at -3,

y=(x-(-3))(x-n)

y=(x+3)(x-n)

subsitute the other point (1,6) to find n

6=(1+3)(1-n)

6=(4)(1-n)

6=4-4n

2=-4n

n=-1/2

so the equation is

$y=(x+3)(x-(-\frac{1}{2}))$ or

$y=(x+3)(x+\frac{1}{2})$

the only problem is that in the picture, the graph of B crosses the x axis at a positive and negative value but the equation I have crosses the x axis at 2 negative values

but without further information, I can't solve for B