(a) y-intercept and slope:
In a linear equation y=b_0+b_1xZSK, the constant b1 be the slope and b0 be the y-intercept form and x is the independent variable and y is the dependent variable.
Comparing the given equation with the general form of linear equation the slope of the equation is 0.5 and the y-intercept is –2.
Therefore, the y-intercept is –2 and the slope of the linear equation (b_1ZSK) is 0.5.
(b) It is known that, the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is upward if b_1 > 0ZSK, the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is downward if \(\displaystyle{b}_{{1}}{<}{0}\)</span>, and the slope of the linear equation \(\displaystyle{y}={b}_{{0}}+{b}_{{1}}{x}\) is horizontal if \(\displaystyle{b}_{{1}}={0}\) .

Thus, in the given equation \(\displaystyle{y}={0.5}{x}–{2},{b}_{{1}}\) is 0.5, which is greater than 0.

Thus, the slope is upward.

(c) Graph by using two points:

The two points \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}},{y}_{{2}}\right)}\) on the given line are obtained as follows:

If x=0

\(\displaystyle{y}={\left({0},{5}\times{0}\right)}-{2}\)

y=-2

Thus, one point on the line is \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},-{2}\right)}\)

If x=4,

\(\displaystyle{y}={\left({0.5}\times{4}\right)}-{2}\)

y=0

Thus, the second point on the line is \(\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({4},{0}\right)}\)

Thus, in the given equation \(\displaystyle{y}={0.5}{x}–{2},{b}_{{1}}\) is 0.5, which is greater than 0.

Thus, the slope is upward.

(c) Graph by using two points:

The two points \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}},{y}_{{2}}\right)}\) on the given line are obtained as follows:

If x=0

\(\displaystyle{y}={\left({0},{5}\times{0}\right)}-{2}\)

y=-2

Thus, one point on the line is \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},-{2}\right)}\)

If x=4,

\(\displaystyle{y}={\left({0.5}\times{4}\right)}-{2}\)

y=0

Thus, the second point on the line is \(\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left({4},{0}\right)}\)