Second degree polynomial regression in R: How to get X given Y?

I do not know of any function that is ready to do this, however this problem can be treated as an optimization problem.

We want to find x in a given interval that will minimize a function. The function I want to minimize is:

objetivo <- function(x, k, model){
  df <- data.frame(x = x)
  (k - predict(model, df))^2
}

Given a y = k value, I want to find a x value that most closely matches the prediction of the template of this k value. In the background this means: find me x that when I apply the template I will get closer to the y that I am looking for .

Example:

Suppose the following data:

x <- runif(100)
y <- 10 + x + x^2 + rnorm(n = 100, mean = 0, sd = 0.1)
plot(x, y)

model<-lm(formula=y~x+I(x^2))Call:lm(formula=y~x+I(x^2))Residuals:Min1QMedian3QMax-0.208921-0.0645060.0015370.0611070.276347Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)9.988180.03736267.335<2e-16***x1.059120.156976.7471.10e-09***I(x^2)0.958220.142296.7341.17e-09***---Signif.codes:0‘***’0.001‘**’0.01‘*’0.05‘.’0.1‘’1Residualstandarderror:0.09685on97degreesoffreedomMultipleR-squared:0.9728,AdjustedR-squared:0.9723F-statistic:1735on2and97DF,p-value:<2.2e-16

Now,usingtheoptimfunctionwecangetwhatwewant.

v<-optim(0,objetivo,k=11,model=model,method="Brent", lower = min(x), upper = max(x))
v$par
[1] 0.6240072

So, we get the value of x which is most likely given y = k = 11 .

Note : The optim function by default uses the Nelder-Mead method, however it releases a warning when I used it in this example. So I switched to the "Brent" method that requires minimum and maximum values for the estimated value of x . This may be good, since by estimating polynomials, it is possible that there are other values that minimize this function.

Adapting the previous answers with data more like the one of the question and understanding that the result of x must be in the 7 categories ("levels") of described values:

v <- c(25,50,75,100,125)
valores <- sort(rep.int(v,6))

set.seed(13)
x <- sample(valores, 36, replace = T)
y <- -26500 + 97000*x + -14*x^2 + rnorm(n = 36, mean = 0, sd = 100000)
plot(x, y, pch = 21)

The Model can look like this:

model <-lm(formula = y ~ x + I(x^2))

As well as the objective function to find the predicted value:

objetivo <- function(x, k, model){
  df <- data.frame(x = x)
  (k - predict(model, df))^2
}

This way by looping with 'optim':

    aux_1 <- numeric(length(y))

    for(i in seq_along(y)){ 
    aux_1[i] <- optim(0, objetivo, k = y[i], model = model, method = "Brent",
 lower = min(x), upper = max(x))$par
    }

    # Tomando os resultados mais próximos:

    result <- round(aux_1,0)
    result
[1] 100  51  50  27 125  25  76 100 124  25  99 124 125  76  75  50  50  74
[19] 125 100  25  74 100  77  26 102  25  74  49  98  75  76 125 125  75  25

If you want only the values present in (25, 50, 75, 100, 125) you can do the function below:

extract <- function(x,y){
y <- sort(y, decreasing = T)

for(i in seq_along(y)){
x[x/y[i] < 1.25 & x/y[i] >= 0.98] = y[i]}
return(x)
}

result <- extract(result,v)
result
[1] 100  50  50  25 125  25  75 100 125  25 100 125 125  75  75  50  50  75
[19] 125 100  25  75 100  75  25 100  25  75  50 100  75  75 125 125  75  25

For values of 'x' closer to each other it may be necessary to further refine the function that looks for the nearest values.

EDIT_1 (11/27/17, 23:35): I was generating an aux_2 vector that was unnecessary in the final published version.

EDIT_2 (06/15/18, 20:20): I have refined the vector of values to be more similar to the data.frame shown as an example and added the good suggestions of Rui Barradas's comment below. Also, since this response lacked approximation to be only with vector values (25, 50, 75, 100, 125), I created the 'extract' function. However, it is a specific function for the values of this example, you can adjust them for more general cases.