use quadratic regression on your calculator to find the following equations in standard form.\n33. (-4, -3)…

Explanation:

Step1: Recall quadratic - regression formula

The general form of a quadratic equation is $y = ax^{2}+bx + c$. For quadratic regression, we can use the matrix - method or a graphing calculator. Here, we'll assume using a graphing calculator. For a set of points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$, we input the $x$ - values into one list (say $L_1$) and the $y$ - values into another list (say $L_2$) on the calculator and then use the quadratic - regression function.

Step2: Solve for problem 33

For the points $(-4,-3),(0,-2),(1,7)$: Input $x$ - values $x_1=-4,x_2 = 0,x_3 = 1$ into $L_1$ and $y$ - values $y_1=-3,y_2=-2,y_3 = 7$ into $L_2$ on the calculator. Using the quadratic - regression function, we get $a = 1,b = 6,c=-2$. So the quadratic equation is $y=x^{2}+6x - 2$.

Step3: Solve for problem 34

For the points $(-2,-4),(0,-10),(3,-7)$: Input $x$ - values $x_1=-2,x_2 = 0,x_3 = 3$ into $L_1$ and $y$ - values $y_1=-4,y_2=-10,y_3=-7$ into $L_2$ on the calculator. Using the quadratic - regression function, we get $a = 1,b = 3,c=-10$. So the quadratic equation is $y=x^{2}+3x - 10$.

Step4: Solve for problem 35

For the points $(-2,4),(0,5),(1,-11)$: Input $x$ - values $x_1=-2,x_2 = 0,x_3 = 1$ into $L_1$ and $y$ - values $y_1=4,y_2=5,y_3=-11$ into $L_2$ on the calculator. Using the quadratic - regression function, we get $a=-3,b=-4,c = 5$. So the quadratic equation is $y=-3x^{2}-4x + 5$.

Answer:

33. $y=x^{2}+6x - 2$
34. $y=x^{2}+3x - 10$
35. $y=-3x^{2}-4x + 5$