Question Video: Finding the Area of the Triangle Whose Vertices Are the Vertex and the Real Roots of a Quadratic Function given Its Graph Mathematics

The figure below represents the function 𝑓(𝑥) = 𝑥² + 𝑚. Find the area of triangle 𝐴𝐵𝐶 given 𝑂𝐴 = 9.


Video Transcript

The figure below represents the function 𝑓 of 𝑥 equals 𝑥 squared plus 𝑚. Find the area of triangle 𝐴𝐵𝐶 given 𝑂𝐴 equals nine.

Let’s begin by sketching in the triangle whose area we’ve been asked to find. It’s this triangle here. We know that the formula for finding the area of a triangle is a half its base multiplied by its perpendicular height. On our diagram, the base of the triangle is the length of the line 𝐵𝐶. And the perpendicular height is the length of the line 𝑂𝐴. So, the area is equal to 𝐵𝐶 multiplied by 𝑂𝐴 over two.

Now, looking at the figure, we can see that there are no numbers on either the 𝑥- or 𝑦-axes, so we haven’t been given the scales to use in determining these lengths. And we must remember that the scales on the two axes could be different. We’re given in the question, though, that 𝑂𝐴 is nine. So, we do have one of the lengths we need. And in fact, we can work out that, on the vertical scale, one small square represents one unit. But this doesn’t automatically mean that the scale on the 𝑥-axis is the same.

What we do know, though, is that if the length of 𝑂𝐴 is nine, then the coordinates of the point 𝐴 will be zero, negative nine because 𝐴 is nine units below the origin. We can use this to determine the value of 𝑚 in the definition of our function 𝑓 of 𝑥. We have 𝑓 of 𝑥 equals 𝑥 squared plus 𝑚. And so, 𝑚, the constant term in this quadratic function, will be the value of the function’s 𝑦-intercept. We can deduce then that our function 𝑓 of 𝑥 is equal to 𝑥 squared minus nine. In fact, it’s just a vertical translation of the graph of 𝑥 squared by nine units downwards.

In order to determine the length of 𝐵𝐶, we need to know the coordinates of the points where this curve intersects the 𝑥-axis, that is, the roots or zeros of this function. We know that at every point on the 𝑥-axis, the value of the function 𝑓 of 𝑥 is equal to zero. So, we can determine these 𝑥-values by letting 𝑓 of 𝑥 equal zero and then solving the resulting equation. We have zero equals 𝑥 squared minus nine. Adding nine to each side, we find that nine is equal to 𝑥 squared. Or we can write this is 𝑥 squared equals nine, if we prefer.

To find the values of 𝑥 then, we take the square root of each side of this equation, giving 𝑥 equals plus or minus the square root of nine. Nine is, of course, a square number. And its square root is three. So, we have that 𝑥 is equal to positive or negative three. From the graph, we can see that the 𝑥-coordinate at 𝐵 must therefore be three and the 𝑥-coordinate at 𝐶 must be negative three. The length of 𝐵𝐶, that’s the difference between three and negative three, is therefore six.

So, having determined both the length of 𝑂𝐴 and 𝐵𝐶, we can substitute into our area calculation, giving six multiplied by nine over two. We can then cancel a factor of two from the numerator and denominator, giving a simplified calculation of three multiplied by nine over one. That’s 27 over one, which is simply 27. The units for this will just be general square units. So, by first finding the 𝑦-intercept of this graph and using it to determine the value of 𝑚 in our function 𝑓 of 𝑥. And then solving the equation 𝑓 of 𝑥 equals zero to find the roots or zeros of this quadratic, which give the 𝑥-coordinates at the points 𝐵 and 𝐶. We found that the area of triangle 𝐴𝐵𝐶 is 27.
