Without Graphing

Determine, WITHOUT GRAPHING, whether the two given quadratic functions have a maximum value or a minimum value and then find the value.

NOTE: What exactly is meant by a max and min value here?

(1) f(x) = -2x^2 + 8x + 3

(2) f(x) = 4x^2 - 8x + 3

Quote:

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Originally Posted by symmetry

Determine, WITHOUT GRAPHING, whether the two given quadratic functions have a maximum value or a minimum value and then find the value.

NOTE: What exactly is meant by a max and min value here?

(1) f(x) = -2x^2 + 8x + 3

(2) f(x) = 4x^2 - 8x + 3

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1.) f'(x) = -4x + 8

0 = -4x + 8;

Solve for x;

x = 2

Thus, the critical point is at x = 2, and that's either a min or a max.

-2(2)^2 + 8(2) + 3

-2*4 + 16 + 3

-8 + 16 + 3

8 + 3 = 11

And thus, it'll be at (2, 11).

We can check by seeing whether the second deriv is pos or neg, and thus determine concavity.

f''(x) = -4, and thus it is a max, since it's concave down.

Therefore, we have a maximum at (2, 11).

Quote:

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Originally Posted by symmetry

(2) f(x) = 4x^2 - 8x + 3

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Follow what I did in 1.

f'(x) = 8x - 8

0 = 8x - 8

x = 1

4(1)^2 - 8(1) + 3

4 - 8 + 3 = -1

f''(x) = 8 (concave up, and thus a min.)

Thus, we have a min. at (1, -1).

Quote:

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Originally Posted by symmetry

NOTE: What exactly is meant by a max and min value here?

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Just that...the lowest point of the graph is obviously the min and the max is the highest point of the graph. Be aware that there is "local" and "global" min/max.

Now, this question came from a precalculus chapter in my state exam prep book. Can you show me how to answer the question WITHOUT using calculus concepts?

I am not into calculus just yet. I am preparing for my state exam and calculus 1 and 2 are the last 4 chapters of my state study book.

Now, how do I answer such a question WITHOUT calculus?

Thanks!