Answer:
This is the same as writing y = sqrt(4(x+5)) - 1
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Step-by-step explanation:
The given graph appears to be a square root function.
The marked points on the curve are:
Reflect those points over the line y = x. This will have us swap the x and y coordinates.
- (-4,1) becomes (1,-4)
- (-1,3) becomes (3,-1)
- (4,5) becomes (5,4)
Recall the process of reflecting over y = x means we're looking at the inverse. The inverse of a square root function is a quadratic.
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Let's find the quadratic curve that passes through (1,-4), (3,-1) and (5,4).
Plug the coordinates of each point into the template y = ax^2+bx+c.
For instance, plug in x = 1 and y = -4 to get...
y = ax^2+bx+c
-4 = a*1^2+b*1+c
-4 = a+b+c
Do the same for (3,-1) and you should get the equation -1 = 9a+3b+c
Repeat for (5,4) and you should get 4 = 25a+5b+c
We have this system of equations
- -4 = a+b+c
- -1 = 9a+3b+c
- 4 = 25a+5b+c
Use substitution, elimination, or a matrix to solve that system. I'll skip steps, but you should get (a,b,c) = (1/4, 1/2, -19/4) as the solution to that system.
In other words
a = 1/4, b = 1/2, c = -19/4
We go from y = ax^2+bx+c to y = (1/4)x^2+(1/2)x-19/4
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Next we complete the square
y = (1/4)x^2+(1/2)x-19/4
y = (1/4)( x^2+2x )-19/4
y = (1/4)( x^2+2x+0 )-19/4
y = (1/4)( x^2+2x+1-1 )-19/4
y = (1/4)( (x^2+2x+1)-1 )-19/4
y = (1/4)( (x+1)^2-1 )-19/4
y = (1/4)(x+1)^2- 1/4 - 19/4
y = (1/4)(x+1)^2 + (-1-19)/4
y = (1/4)(x+1)^2 - 20/4
y = (1/4)(x+1)^2 - 5
The equation is in vertex form with (-1,-5) as the vertex. It's the lowest point on this parabola. Placing it into vertex form allows us to find the inverse fairly quickly.
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The last batch of steps is to find the inverse.
Swap x and y. Then solve for y.
y = (1/4)(x+1)^2 - 5
x = (1/4)(y+1)^2 - 5
x+5 = (1/4)(y+1)^2
(1/4)(y+1)^2 = x+5
(y+1)^2 = 4(x+5)
y+1 = sqrt(4(x+5))
y = sqrt(4(x+5)) - 1
I'll let the student check each point to confirm they are on the curve y = sqrt(4(x+5)) - 1.
You can also use a tool like GeoGebra to verify the answer.