Question

svlc algebra 1a - standard (15260)
linear functions
which linear function is represented by the graph?
$f(x)=\frac{1}{2}x + 1$
$f(x)=-\frac{1}{2}x + 1$
$f(x)=-2x + 1$

Explanation:

Step1: Recall slope-intercept form

The slope - intercept form of a linear function is \(f(x)=mx + b\), where \(m\) is the slope and \(b\) is the \(y\) - intercept. From the graph, we can see that the line crosses the \(y\) - axis at \((0,1)\), so \(b = 1\).

Step2: Calculate the slope \(m\)

The formula for the slope \(m\) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's use the points \((0,1)\) and \((4,-1)\). Here, \(x_1 = 0,y_1 = 1,x_2 = 4,y_2=-1\).
Substitute into the slope formula: \(m=\frac{-1 - 1}{4-0}=\frac{-2}{4}=-\frac{1}{2}\).

Step3: Write the linear function

Since \(m =-\frac{1}{2}\) and \(b = 1\), the linear function is \(f(x)=-\frac{1}{2}x + 1\). We can also verify by plugging in the point \((-4,3)\) into the function:
Left - hand side: \(f(-4)\)
Right - hand side: \(-\frac{1}{2}\times(-4)+1=2 + 1=3\), which matches the \(y\) - value of the point \((-4,3)\).
If we check the first function \(f(x)=\frac{1}{2}x + 1\), when \(x=-4\), \(f(-4)=\frac{1}{2}\times(-4)+1=-2 + 1=-1
eq3\).
For the third function \(f(x)=-2x + 1\), when \(x = - 4\), \(f(-4)=-2\times(-4)+1=8 + 1=9
eq3\).

Answer:

\(f(x)=-\frac{1}{2}x + 1\) (the second option)