For the quadratic equation y=(x-3)^2-2. Determine: A) whether the graph opens up or down; B) the y-intercept; C)the x-intercepts; D) the points of the vertex. and D) graph the equation with at lease 10 pairs of points.
Vertex Form: Y = a(x-h)^2 + k.
A. The parabola opens upward because "a"
is positive.
B. Y = (x-3)^2 - 2.
Let X = o and solve for Y:
Y = (0-3)^2 - 2 = 7. = Y-int.
C. Let Y = 0 and solve forX:
Y = (x-3)^2-2 = 0.
(x-3)^2 = 2
Take sgrt of both sides:
x-3 = +-sqrt2
X = +-1.414 + 3 = 4.14, and 1.59. = X-intercepts.
D. V(h,k) = V(3,-2).
E. Select values of X below and above
h and calculate the corresponding value of Y.
(x,y). Y = (x-3)^2 - 2.
(-2,23)
(-1,14)
(0,7)
(1,2)
(2,-1)
V(3,-2)
(4,-1)
(5,2)
(6,7)
(7,14)
(8,23).
for the quadratic equation y=(x-3)^2-2 determine A) weather the graph opens up or down B) the y- intercept; C) the x intercept D) the points of the vertex and E ) graph the equation with at least 10 pairs of points
To answer these questions about the quadratic equation y=(x-3)^2-2, we will go through each one step by step.
A) To determine whether the graph opens up or down, we can look at the coefficient of the x^2 term. In this case, the coefficient is positive, which means the graph will open upward.
B) To find the y-intercept, we need to substitute x=0 into the equation and solve for y. Plugging x=0 into the equation, we get y = (0-3)^2-2 = 9-2 = 7. Therefore, the y-intercept is at the point (0, 7).
C) To find the x-intercepts, we need to set y=0 and solve for x. Setting y=0 in the equation, we have (x-3)^2 - 2 = 0. Now let's solve for x:
(x-3)^2 - 2 = 0
(x-3)^2 = 2
x-3 = ±√2
x = 3 ± √2
So, the x-intercepts are at the points (3 + √2, 0) and (3 - √2, 0).
D) To find the coordinates of the vertex, we observe that the vertex of a quadratic equation in vertex form y = a(x-h)^2 + k is at the point (h, k). In this case, the vertex form of the equation is y = (x-3)^2 - 2. Therefore, the vertex is located at the point (3, -2).
D) Finally, let's graph the equation with at least 10 pairs of points. We already know the following points: y-intercept (0, 7), x-intercepts (3 + √2, 0) and (3 - √2, 0), and vertex (3, -2). You can choose any other x values and find the corresponding y values to get more points. For example:
When x = 1, y = (1-3)^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2. So we have the point (1, 2).
When x = 2, y = (2-3)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1. So we have the point (2, -1).
When x = 4, y = (4-3)^2 - 2 = (1)^2 - 2 = 1 - 2 = -1. So we have the point (4, -1).
When x = 5, y = (5-3)^2 - 2 = (2)^2 - 2 = 4 - 2 = 2. So we have the point (5, 2).
...
By plotting these points and connecting them, you can create the graph of the quadratic equation y=(x-3)^2-2.