The value of m is maximized when the axis of symmetry of the graph y=f(x) is the straight line x = ?

[satsuimiamo]

from the 2017 EJU. im having trouble with this question in general. i can do 1-2, but im not sure what 3 and 4 are asking. when i use the answers for 2 to plug in for 3 and 4, i dont get the correct answers. when graphed, i can see the maximum and minimum of the function, but i dont know how someone would get there without using a desmos slider. (answers are here, scroll down to the math section. its Q1 for both courses)

 

[Dr.Peterson]

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from the 2017 EJU. im having trouble with this question in general. i can do 1-2, but im not sure what 3 and 4 are asking. when i use the answers for 2 to plug in for 3 and 4, i dont get the correct answers. when graphed, i can see the maximum and minimum of the function, but i dont know how someone would get there without using a desmos slider. (answers are here, scroll down to the math section. its Q1 for both courses)

You found that m = -1/8 a^2 - 1, right?

Use the same reasoning you used to answer part (1) to find the maximum of this function (or just look at the equation and know). That will give you the value of a that yields the largest possible value of the minimum, m. Put that value of a into the given function definition, and find its axis of symmetry.

 

[satsuimiamo]

You found that m = -1/8 a^2 - 1, right?

Use the same reasoning you used to answer part (1) to find the maximum of this function (or just look at the equation and know). That will give you the value of a that yields the largest possible value of the minimum, m. Put that value of a into the given function definition, and find its axis of symmetry.

do i use m to find it? but im not sure how to find the value for a that would give me the maximum. thank you!

 

[Dr.Peterson]

do i use m to find it? but im not sure how to find the value for a that would give me the maximum. thank you!

Are you allowing the name of a variable to influence you?

What if you were asked to find the vertex of y = -1/8 x^2 - 1, rather than the maximum of m = -1/8 a^2 - 1?

 

[satsuimiamo]

Are you allowing the name of a variable to influence you?

What if you were asked to find the vertex of y = -1/8 x^2 - 1, rather than the maximum of m = -1/8 a^2 - 1?

i think i understand now! thank you so much. but for 4), i was actually able to just work out that plugging in 4 would give me a lesser value than -2, but is there a more efficient way to figure out the minimum?

 

[Dr.Peterson]

i think i understand now! thank you so much. but for 4), i was actually able to just work out that plugging in 4 would give me a lesser value than -2, but is there a more efficient way to figure out the minimum?

No, that's just what you have to do. Since the function is concave-down, the absolute minimum is at an endpoint of the domain, which you found in part 2; so you have to test both of them.