The parametric equation {x=t^2+4t+8 ; y=t+4
a. Reduce the equation to simplest rectangular form
b. Determine the Center and the type of graph of the resulting equation.
c. Graph the equation in rectangular form
polar can get kind of complicated, but this one's a piece of cake:
y = t+4, so t = y-4
x = (y-4)^2 + 4(y-4) + 8
Now expand it out and note that it's just a parabola.
To simplify the parametric equation {x=t^2+4t+8 ; y=t+4} into rectangular form, we need to eliminate the parameter (t) by expressing one variable in terms of the other. Let's start with the equation for x.
a. To eliminate t from the equation for x, we can use the quadratic formula. Rearrange the equation x = t^2 + 4t + 8 to form a quadratic equation: t^2 + 4t + (8 - x) = 0.
Now, apply the quadratic formula: t = (-4 ± √(4^2 - 4(1)(8 - x))) / (2(1)).
Simplify it further: t = (-4 ± √(16 - 32 + 4x)) / 2.
Combine the like terms inside the square root: t = (-4 ± √(4x - 16)) / 2.
Simplify the expression: t = -2 ± √(x - 4).
So, we have t = -2 + √(x - 4) and t = -2 - √(x - 4).
b. Now, let's determine the center and type of graph for the resulting equation.
The equation y = t + 4 tells us that for every given value of t, y increases by 1. Therefore, the graph will be a straight line.
The center of the graph can be determined by evaluating the values of x and y at a certain value of t. Let's find the center by substituting t = 0 into the parametric equations:
x = (0)^2 + 4(0) + 8 = 8,
y = 0 + 4 = 4.
So, the center of the graph is (8, 4).
c. To graph the equation in rectangular form, substitute the expression for x in terms of y:
x = (-2 + √(y - 4))^2 + 4(-2 + √(y - 4)) + 8
Simplify and rearrange to put the equation in a standard form if needed.
Plotting several points on the graph by substituting different values of y will allow you to see the shape of the graph. Additionally, you can check for symmetry and key points to get a clearer picture.
I hope this helps! Let me know if you have any further questions.