# The graph of which function has a minimum located at (4, -3)?

f(x) = -1/2x^{2} + 4x - 11

f(x) = -2x^{2} + 16x - 35

f(x) =1/2x^{2} - 4x + 5

f(x) = 2x^{2} - 16x + 35

**Solution:**

Given minimum at (4, -3)

Minimum occurs when the second derivative of the function is positive.

Let us check all the options

1. f(x) = -1/2x^{2} + 4x - 11 : f’(x) = -x + 4: f’’(x) = -1(negative)

2. f(x) = -2x^{2} + 16x - 35 : f’(x) = -4x + 16: f’’(x) = -4(negative)

3. f(x) = 1/2x^{2} - 4x + 5 : f’(x) = x - 4: f’’(x) = 1(positive)

4. f(x) = 2x^{2} - 16x + 35 : f’(x) = 4x - 16: f’’(x) = 4(positive)

Now, we have two suitable functions, given that it is located at (4, -3)

So, put x = 4 and y = -3

Check if f(x) = y in both the functions.

f(x) = 1/2 x^{2} - 4x + 5

= 1/2 (4)^{2} - 4(4) + 5

= 8 - 16 + 5

= -3 = y

f(x) =y= 2x^{2} - 16x + 35

= 2(4)^{2}-16(4)+35

=32-64+35

=3 ≠ -3

Therefore, f(x) =1/2x^{2} - 4x + 5 is the required equation.

## The graph of which function has a minimum located at (4, -3)?

**Summary:**

The graph of the function which has a minimum located at (4, -3) is f(x) = 1/2x^{2} - 4x + 5.

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