## Adi3 one year ago for each of the following quadratic functions: (i) Find the coordinates for the vertex (ii) determine whether the vertex is max or min (iii) find the range for the functions (a) f(x) = 2x^2 +4

@SolomonZelman

@AaronAndyson

@imqwerty

4. SolomonZelman

$$\color{black}{ \displaystyle f(x)=2x^2+4~~~~\Longrightarrow ~~~~f(x)=2(x-0)^2+4 }$$ And this is in the form of $$\color{black}{ \displaystyle f(x)=2(x-h)^2+k }$$ (with vertex $$(h,k)$$ )

ok so the vertex is 0,4

6. SolomonZelman

yes, correct.

The vertex is min

8. SolomonZelman

If the leading coefficient is POSITIVE, then parabola opens UP, the vertex is the MINIMUM point (or the absolute MINIMUM). If the leading coefficient is NEGATIVE, then parabola opens DOWN, the vertex is the MAXIMUM point (or the absolute MAXIMUM).

9. SolomonZelman

Yes, so the vertex is minimum.

but you said max

11. SolomonZelman

When the leading coefficient is negative.

ohh ok

13. SolomonZelman

But, as I said, when leading coefficient is positive (in your case it is 2), then the vertex is minimum.

14. SolomonZelman

And the range of the opening up parabola will start from the vertex, but has no restrictions on how large the output can be....

so it is (y:y≤4)

16. SolomonZelman

Yes, that is exactly right.

??

18. SolomonZelman

Or, in interval notation $${\rm y} \in [4,+\infty)$$

we did not learn that yet.

20. SolomonZelman

All that means is that the range goes from and including 4, and ends at positive infinity (meaning that it goes up forever endlessly).

21. SolomonZelman

don't want to throw in confusion.... in any case you are right y$$\le$$4.

22. SolomonZelman

Any questions about what we have done?

nop, thanks a lot really apreciate, really needed it, i have a quiz tmrw.

24. SolomonZelman

Good luck on the quiz:)

thanks mate

thanks @imqwerty for seeing my question.

27. imqwerty

lolol XD