# Without Graphing

• Jan 24th 2007, 12:45 PM
symmetry
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
• Jan 24th 2007, 12:54 PM
AfterShock
Quote:

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

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).
• Jan 24th 2007, 12:57 PM
AfterShock
Quote:

Originally Posted by symmetry

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

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).
• Jan 24th 2007, 01:00 PM
AfterShock
Quote:

Originally Posted by symmetry

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

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.
• Jan 24th 2007, 05:13 PM
symmetry
ok
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!