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20-90x-50x^2
find the vertex

User Li Dong
by
5.4k points

2 Answers

4 votes

20-90x-50x^2, written in standard form, would be -50x^2 - 90x + 20. Let's use the method of completing the square to determine the vertex:

Divide all terms in -50x^2 - 90x + 20 by -50, obtaining -50(x^2 + (9/5)x - 2/5)

The coefficient of x is 9/5. Divide this by 2, obtaining 9/10, and then square this result: (9/10)^2 = 81/100. Add this to x^2 + (9/5)x - 2/5 and then subtract it, resulting in x^2 + (9/5)x - 2/5 + 81/100 - 81/100, or

x^2 + (9/5)x + 81/100 - 2/5 - 81/100, or

(x + 9/10)^2 - 12100

Then the vertex location is (-9/10, -12100).

A faster approach would be to use x = -b / (2a) to determine the x-coordinate of the vertex and then to evaluate the function at this x-value to determine the y-coordinate. Here x = -(-90) / (-100) = -9/10 (same as before).

User Audun Kjelstrup
by
5.2k points
2 votes
I think it’s your answer
20-90x-50x^2 find the vertex-example-1
User Antoine Pietri
by
5.2k points