Finding the range of y=x2+2x+42x2+4x+9 (and y=quadraticquadratic in general)

Kinsley Moon

Kinsley Moon

Answered

2022-01-31

Finding the range of y=x2+2x+42x2+4x+9 (and y=quadraticquadratic in general)

Answer & Explanation

Sean Becker

Sean Becker

Expert

2022-02-01Added 16 answers

The question can be easily solved by this technique: As y=x2+2x+42x2+4x+92y=2x2+4x+912x2+4x+9. Thus, 2y=112(x+1)2+7 Squares can never be less than zero so the minmum value of the function: 2(x+1)2+7 would be 7, or Maximum value of 12(x+1)2+7 is 17. This tells that miminum value of y will be 37. From here you can easily tell the maximum and minimum values: y[37,12)
ebbonxah

ebbonxah

Expert

2022-02-02Added 15 answers

y=x2+2x+42x2+4x+9=2x2+4x+82(2x2+4x+9)=2x2+4x+912(2x2+4x+9)=12(112x2+4x+9)2y=112x2+4x+9.
Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by 2 to get the required extremes(taking half since we get values for 2y and not y).

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?