If f(x)=x^2-1 and g(x)=3f(x),graph g(x) and write the equation for g(x)
To graph g(x), we need to determine the equation for g(x) by substituting f(x) into g(x).
Given that f(x) = x^2 - 1, we can find g(x) by multiplying f(x) by 3.
g(x) = 3 * f(x)
g(x) = 3 * (x^2 - 1)
g(x) = 3x^2 - 3
Now, let's proceed to graph g(x):
The graph of g(x) will be a parabola because it is a quadratic function. We can easily find the vertex, y-intercept, and the direction of the opening of the parabola by analyzing the equation.
The coefficient of x^2 is positive, so the parabola opens upward. The vertex can be found using the formula x = -b/(2a):
x = -(-3)/(2*3)
x = 1/2
To find the y-coordinate of the vertex, we substitute this value of x into the equation for g(x):
g(1/2) = 3(1/2)^2 - 3
g(1/2) = 3/4 - 3
g(1/2) = -8/4
g(1/2) = -2
Therefore, the vertex of the parabola is (1/2, -2).
Now, let's find the y-intercept by plugging in x = 0 into the equation for g(x):
g(0) = 3(0)^2 - 3
g(0) = -3
The y-intercept is (0, -3).
We can also find the x-intercepts by setting g(x) = 0:
0 = 3x^2 - 3
3x^2 = 3
x^2 = 1
x = ±1
The x-intercepts are (-1, 0) and (1, 0).
Now, let's plot these points and sketch the graph of g(x).
The vertex is (1/2, -2), the y-intercept is (0, -3), and the x-intercepts are (-1, 0) and (1, 0). The graph will be a parabola opening upward, which passes through these points.
The graph of g(x) should resemble a "U" shape opening upward, with the vertex at (1/2, -2), horizontal axis of symmetry at x = 1/2, and y-intercept at (0, -3).
I apologize, but I am unable to draw the graph here. Please refer to a graphing tool or software to visualize the graph of g(x).