### Video Transcript

The figure below represents the function π of π₯ equals π₯ squared plus π. Find the area of triangle π΄π΅πΆ given ππ΄ equals nine.

Letβs begin by sketching in the triangle whose area weβve been asked to find. Itβs 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 π΅πΆ. And the perpendicular height is the length of the line ππ΄. So, the area is equal to π΅πΆ multiplied by ππ΄ over two.

Now, looking at the figure, we can see that there are no numbers on either the π₯- or π¦-axes, so we havenβt been given the scales to use in determining these lengths. And we must remember that the scales on the two axes could be different. Weβre given in the question, though, that ππ΄ 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 doesnβt automatically mean that the scale on the π₯-axis is the same.

What we do know, though, is that if the length of ππ΄ is nine, then the coordinates of the point π΄ will be zero, negative nine because π΄ is nine units below the origin. We can use this to determine the value of π in the definition of our function π of π₯. We have π of π₯ equals π₯ squared plus π. And so, π, the constant term in this quadratic function, will be the value of the functionβs π¦-intercept. We can deduce then that our function π of π₯ is equal to π₯ squared minus nine. In fact, itβs just a vertical translation of the graph of π₯ squared by nine units downwards.

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

To find the values of π₯ then, we take the square root of each side of this equation, giving π₯ 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 π₯ is equal to positive or negative three. From the graph, we can see that the π₯-coordinate at π΅ must therefore be three and the π₯-coordinate at πΆ must be negative three. The length of π΅πΆ, thatβs the difference between three and negative three, is therefore six.

So, having determined both the length of ππ΄ and π΅πΆ, 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. Thatβs 27 over one, which is simply 27. The units for this will just be general square units. So, by first finding the π¦-intercept of this graph and using it to determine the value of π in our function π of π₯. And then solving the equation π of π₯ equals zero to find the roots or zeros of this quadratic, which give the π₯-coordinates at the points π΅ and πΆ. We found that the area of triangle π΄π΅πΆ is 27.