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algebra: the role of slope and intercepts (part 2) ex 1: use the graph …

Question

algebra: the role of slope and intercepts (part 2)
ex 1: use the graph of the line shown to answer the following:
a) what is the y - intercept?
b) what is the x - intercept?
c) is this a partial or direct variation?
d) what is the slope? (use a rate triangle)
e) write the equation of the line.
ex 2: graph each of the following lines using a table of values.
a) ( y = 3x - 4 )
b) ( y = -\frac{1}{2}x + 1 )

Explanation:

Ex 1:

a)

The \(y -\)intercept is the point where the line crosses the \(y -\)axis. Looking at the graph, when \(x = 0\), \(y=2\). So the \(y -\)intercept is \(2\).

b)

The \(x -\)intercept is the point where the line crosses the \(x -\)axis. Looking at the graph, when \(y = 0\), \(x=-2\). So the \(x -\)intercept is \(-2\).

c)

The equation of a direct variation is \(y=kx\) (\(b = 0\) in \(y=kx + b\)). The equation of a partial variation is \(y=kx + b\) (\(b
eq0\)). Since the \(y -\)intercept \(b = 2
eq0\), this is a partial variation.

d)

Using the rate - triangle (rise over run). We can take two points on the line, say \((-2,0)\) and \((0,2)\). The rise (\(\Delta y\)) is \(2-0=2\) and the run (\(\Delta x\)) is \(0-(-2)=2\). The slope \(m=\frac{\Delta y}{\Delta x}=\frac{2}{2}=1\).

e)

The slope - intercept form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the \(y -\)intercept. We found \(m = 1\) and \(b=2\), so the equation of the line is \(y=x + 2\).

Ex 2:

a) For \(y = 3x-4\):

  • When \(x=-2\), \(y=3\times(-2)-4=-6 - 4=-10\)
  • When \(x=-1\), \(y=3\times(-1)-4=-3-4=-7\)
  • When \(x = 0\), \(y=3\times0-4=-4\)
  • When \(x = 1\), \(y=3\times1-4=3 - 4=-1\)
  • When \(x = 2\), \(y=3\times2-4=6 - 4=2\)

b) For \(y=-\frac{1}{2}x + 1\):

  • When \(x=-2\), \(y=-\frac{1}{2}\times(-2)+1=1 + 1=2\)
  • When \(x=-1\), \(y=-\frac{1}{2}\times(-1)+1=\frac{1}{2}+1=\frac{3}{2}\)
  • When \(x = 0\), \(y=-\frac{1}{2}\times0+1=1\)
  • When \(x = 1\), \(y=-\frac{1}{2}\times1+1=-\frac{1}{2}+1=\frac{1}{2}\)
  • When \(x = 2\), \(y=-\frac{1}{2}\times2+1=-1 + 1=0\)

Answer:

Ex 1:

a) \(2\)
b) \(-2\)
c) Partial variation
d) \(1\)
e) \(y=x + 2\)

Ex 2:

a)

\(x\)\(y\)
\(-1\)\(-7\)
\(0\)\(-4\)
\(1\)\(-1\)
\(2\)\(2\)

b)

\(x\)\(y\)
\(-1\)\(\frac{3}{2}\)
\(0\)\(1\)
\(1\)\(\frac{1}{2}\)
\(2\)\(0\)