To sketch the graph of each function, we can start by finding some key points on the graph and then connect them to get a sense of its shape.
1. y = √(x - 1): x ≥ 2
First, let's find the y-intercept by substituting x = 2 into the equation:
y = √(2 - 1) = √1 = 1. So, the graph passes through the point (2, 1).
Next, let's find the x-intercept by setting y to 0:
0 = √(x - 1)
Square both sides: 0 = x - 1
Solve for x: x = 1. So, the graph passes through the point (1, 0).
We can also find a few more points by substituting other values of x:
- When x = 3, y = √(3 - 1) = √2.
- When x = 4, y = √(4 - 1) = √3.
Plot these points on a coordinate plane, and draw a smooth curve following the shape indicated by these points. The graph should start at (1, 0) and extend upwards to the right.
2. f(x) = 3 - x^2: -4 ≤ x ≤ 4
To sketch this quadratic function, we can start by finding the vertex and y-intercept.
To find the y-intercept, set x = 0:
f(0) = 3 - 0^2 = 3. So, the graph passes through (0, 3).
The vertex of a quadratic function is given by the formula x = -b / (2a), where a and b are the coefficients of the quadratic term and linear term, respectively.
In this case, a = -1 and b = 0, so the vertex is at x = -(0) / (2*(-1)) = 0. The y-coordinate of the vertex can be found by substituting x = 0 into the equation:
f(0) = 3 - 0^2 = 3. So, the vertex is (0, 3).
Now, we can plot the vertex (0, 3) and the y-intercept (0, 3) on the coordinate plane. Since the given range is -4 ≤ x ≤ 4, we can determine that the graph is a symmetrical parabola that opens downwards.
To find a few more points, we can substitute a few more values of x:
- When x = -2, f(-2) = 3 - (-2)^2 = 3 - 4 = -1.
- When x = -3, f(-3) = 3 - (-3)^2 = 3 - 9 = -6.
- When x = 2, f(2) = 3 - 2^2 = 3 - 4 = -1.
- When x = 3, f(3) = 3 - 3^2 = 3 - 9 = -6.
Plot these points and draw a smooth curve connecting them. The graph should be a downward-opening parabola symmetric about the line x = 0.
Sketch the graph of the following
y=√(x-1):x≥2
f(x)=3-x^2:-4≤x≤4
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