Find the graph of the following functions.
f\left(x\right)=4x-2
f is a linear function. We need to find two points to determine its graph.
An equation of f is y=4x-2. We plug in 0 for x and obtain:
y=-2
We plug in 0 for y to find x :
0 = 4x- 2
4x = 2
x=\dfrac{1}{2}
The x and y intercepts are:
- \left(0,-2\right)
- \left(\dfrac{1}{2},0\right)
Therefore, the line crosses the points \left(0,-2\right) and \left(\dfrac{1}{2},0\right). The graph of f is:
f\left(x\right)=-x+1
f is a linear function. We need to find two points to determine its graph.
An equation of f is y=-x+1. We plug in 0 for x and obtain:
y=1
We plug in 0 for y to find x :
0=-x+1
x=1
The x and y intercepts are:
- \left(0{,}1\right)
- \left(1{,}0\right)
Therefore, the line crosses the points \left(0{,}1\right) and \left(1{,}0\right). The graph of f is:
f\left(x\right)=2
The equation of f implies that its graph is a straight line parallel to x -axis at a distance of 2 units above the x -axis. The graph of f is:
f\left(x\right)=-2x+3
f is a linear function. We need to find two points to determine its graph.
An equation of f is y=-2x+3. We plug in 0 for x and obtain:
y=3
We plug in 0 for y to find x :
0 = -2x+3
x=\dfrac{3}{2}
The x and y intercepts are:
- \left(0{,}3\right)
- \left(\dfrac{3}{2},0\right)
Therefore, the line crosses the points \left(0{,}3\right) and \left(\dfrac{3}{2},0\right). The graph of f is:
f\left(x\right)=x+1
f is a linear function. We need to find two points to determine its graph.
We plug in 0 for x and obtain:
y=1
We plug in 0 for y to find x :
x=-1
The x and y intercepts are:
- \left(0{,}1\right)
- \left(-1{,}0\right)
Therefore, the line crosses the points \left(0{,}1\right) and \left(-1{,}0\right). The graph of f is:
f\left(x\right)=2x-\dfrac{4}{3}
f is a linear function. We need to find two points to determine its graph.
We plug in 0 for x and obtain:
y=-\dfrac{4}{3}
We plug in 0 for y to find x :
0 = 2x- \dfrac{4}{3}
x = \dfrac{2}{3}
The x and y intercepts are:
- \left(0,-\dfrac{4}{3}\right)
- \left(\dfrac{2}{3},0\right)
Therefore, the line crosses the points \left(0,-\dfrac{4}{3}\right) and \left(\dfrac{2}{3},0\right). The graph of f is:
