Calculate Negative Area: Finding Area Below f(x) = x² + 16

Question

Find the negative area of the function

f(x)=x2+16 f(x)=x^2+16

Video Solution

Solution Steps

00:00 Find the negative domain of the function
00:03 Notice the coefficient of X squared is positive, so the function is happy
00:08 Negative domain is actually below the X-axis
00:15 Therefore substitute Y=0 to find intersection points with X-axis
00:18 Isolate X
00:24 Any number squared will always be positive
00:30 Therefore there are no intersection points with X-axis
00:35 Positive function without intersection points with X-axis
00:39 The positive domain is above the X-axis
00:43 The function is always above X-axis, therefore always positive
00:52 And this is the solution to the question

Step-by-Step Solution

To solve this problem, we'll follow the steps outlined in our analysis.

  • Step 1: Analyze the function's form f(x)=x2+16 f(x) = x^2 + 16 . Here, a=1,b=0, a=1, b=0, and c=16 c=16 .

  • Step 2: Find the vertex to see if the function ever takes negative values. The vertex x x is calculated by x=b2a=0 x = -\frac{b}{2a} = 0 .

  • Step 3: Evaluate f(x) f(x) at this vertex: f(0)=02+16=16 f(0) = 0^2 + 16 = 16 .

  • Step 4: Determine when f(x) < 0 . Since f(x)=x2+1616 f(x) = x^2 + 16 \geq 16 for all real numbers x x , the function is always positive.

  • Step 5: Compare the finding against multiple-choice options. The choice indicating that f(x) f(x) is always positive is the correct one: "Always positive".

The conclusion, therefore, is as follows: the function f(x)=x2+16 f(x) = x^2 + 16 is always positive, and there is no negative area under the graph relative to the x-axis.

Answer

Always positive