I. How long is the lunch period?
1 answer:
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This is an interesting question. Wish there were more questions of this kind.
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol