3x² + 5x - 2 = 0
3x² + 6x - x - 2 = 0
3x(x) + 3x(2) - 1(x) - 1(2) = 0
3x(x + 2) - 1(x + 2) = 0
(3x - 1)(x + 2) = 0
3x - 1 = 0 or x + 2 = 0
+ 1 + 1 - 2 - 2
3x = 1 or x = -2
3 3 1 1
x = ¹/₃ or x = -2
f(x) = 3x² + 5x - 2
f(¹/₃) = 3(¹/₃)² + 5(¹/₃) - 2
f(¹/₃) = 3(¹/₉) + 1²/₃ - 2
f(¹/₃) = ¹/₃ - ¹/₃
f(¹/₃) = 0
(x, f(x)) = (¹/₃, 0)
f(x) = 3x² + 5x - 2
f(-2) = 3(-2)² + 5(-2) - 2
f(-2) = 3(4) - 10 - 2
f(-2) = 12 - 12
f(-2) = 0
(x, f(x)) = (-2, 0)
Vertical Asymptotes: ¹/₃ or -2
Horizontal Asymptotes: 0
Oblique Asymptote: No Asymptotes
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
Answer:
use dat brain
Step-by-step explanation:
Answer:
Y=5x
Step-by-step explanation:
slope equation
<u>y-y1</u>
x-x1
10-0=10
2-0=2
10/2=5
