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# graphing question

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Let f(x) = ((x - 9)^2 - 3)/2.

The equation y=f(x) is graphed, and the x and y intercepts of the graph are connected to form a polygon. What is the area of that polygon?

Jan 29, 2021

#1
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To solve this question, we must first find the intercepts of the curve of the parabola in vertex form: $$f(x)=\frac{1}{2}(x-9)^2-\frac{3}{2}$$.

The y-intercept of a function occurs when x = 0. Let's find the corresponding y-coordinate. Since this equation is a function by nature, we should expect that there exists no more than one y-intercept, if one exists at all.

$$f(0)=\frac{1}{2}(0-9)^2-\frac{3}{2}\\ f(0)=\frac{81}{2}-\frac{3}{2}\\ f(0)=\frac{78}{2}\\ f(0)=39\\ \text{Coordinates of Y-Intercept: (0,39)}$$

The x-intercept of a function occurs when $$f(x)=0$$. Let's find the corresponding x-coordinate(s). Unlike the y-intercept where we could guaratee that only one exists, there may be more than one corresponding x-coordinate that makes the value of the function 0.

$$\frac{1}{2}(x-9)^2-\frac{3}{2}=0\\ \frac{1}{2}(x-9)^2=\frac{3}{2}\\ (x-9)^2=3\\ |x-9|=\sqrt{3}\\ x-9=\sqrt{3} \text{ or } x-9=-\sqrt{3}\\ x=9+\sqrt{3}\text{ or } x=9-\sqrt{3}\\ \text{Coordinates of X-Intercept: }(9-\sqrt{3},0)\text{ and }(9+\sqrt{3},0)$$

Let's graph the equation alongside with the points of the intercepts and connect them with straight lines. Let's see what figure emerges! Pardon my editing, but I did my best to provide an image of this situation:

Let's not lose focus on our goal with this problem. Our final goal is to find the area of the polygon formed by the intercepts of the $$f(x)$$.

$$A_\Delta=\frac{1}{2}bh$$ is the formula for the area of the triangle where b is the length of the base and h is the perpendicular height of the triangle.

$$b = \left(9 + \sqrt{3}\right) - \left(9 - \sqrt{3}\right)\\ b = 2\sqrt{3}\\ h=39\\ A_\Delta = \frac{1}{2}*2\sqrt{3}*39\\ A_\Delta = 39\sqrt{3}$$

Check my work and make sure that I did not make any errors, but 39√3 should be the area of the triangle.

Jan 29, 2021

#1
+1

To solve this question, we must first find the intercepts of the curve of the parabola in vertex form: $$f(x)=\frac{1}{2}(x-9)^2-\frac{3}{2}$$.

The y-intercept of a function occurs when x = 0. Let's find the corresponding y-coordinate. Since this equation is a function by nature, we should expect that there exists no more than one y-intercept, if one exists at all.

$$f(0)=\frac{1}{2}(0-9)^2-\frac{3}{2}\\ f(0)=\frac{81}{2}-\frac{3}{2}\\ f(0)=\frac{78}{2}\\ f(0)=39\\ \text{Coordinates of Y-Intercept: (0,39)}$$

The x-intercept of a function occurs when $$f(x)=0$$. Let's find the corresponding x-coordinate(s). Unlike the y-intercept where we could guaratee that only one exists, there may be more than one corresponding x-coordinate that makes the value of the function 0.

$$\frac{1}{2}(x-9)^2-\frac{3}{2}=0\\ \frac{1}{2}(x-9)^2=\frac{3}{2}\\ (x-9)^2=3\\ |x-9|=\sqrt{3}\\ x-9=\sqrt{3} \text{ or } x-9=-\sqrt{3}\\ x=9+\sqrt{3}\text{ or } x=9-\sqrt{3}\\ \text{Coordinates of X-Intercept: }(9-\sqrt{3},0)\text{ and }(9+\sqrt{3},0)$$

Let's graph the equation alongside with the points of the intercepts and connect them with straight lines. Let's see what figure emerges! Pardon my editing, but I did my best to provide an image of this situation:

Let's not lose focus on our goal with this problem. Our final goal is to find the area of the polygon formed by the intercepts of the $$f(x)$$.

$$A_\Delta=\frac{1}{2}bh$$ is the formula for the area of the triangle where b is the length of the base and h is the perpendicular height of the triangle.

$$b = \left(9 + \sqrt{3}\right) - \left(9 - \sqrt{3}\right)\\ b = 2\sqrt{3}\\ h=39\\ A_\Delta = \frac{1}{2}*2\sqrt{3}*39\\ A_\Delta = 39\sqrt{3}$$

Check my work and make sure that I did not make any errors, but 39√3 should be the area of the triangle.

Guest Jan 29, 2021