Question

If

Answer 1

Answer:

$\large\boxed{Q1.\ q(x)=\dfrac{1}{2}x+\dfrac{7}{2},\ q(2)=\dfrac{9}{2}}\\\boxed{Q2.\ t(x)=\dfrac{1}{2}x+\dfrac{7}{2},\ t(0)=\dfrac{7}{2}}$

Step-by-step explanation:

$\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{We have}\\\\q(-1)=3\to(-1,\ 3)\\q(3)=5\to(3,\ 5)$

$\text{Calculate the slope:}\\\\m=\dfrac{5-3}{3-(-1)}=\dfrac{2}{4}=\dfrac{2:2}{4:2}=\dfrac{1}{2}\\\\\text{We have the equation:}\\\\y=\dfrac{1}{2}x+b\\\\\text{Put the coordinates of the point (-1, 3) to the equation:}\\\\3=\dfrac{1}{2}(-1)+b\\\\3=-\dfrac{1}{2}+b\qquad\text{add}\ \dfrac{1}{2}\ \text{to both sides}\\\\3\dfrac{1}{2}=b\to b=3\dfrac{1}{2}=\dfrac{7}{2}$

$q(x)=\dfrac{1}{2}x+\dfrac{7}{2}\\\\q(2)-\text{put x = 2 to the equation:}\\\\q(2)=\dfrac{1}{2}(2)+\dfrac{7}{2}=\dfrac{2}{2}+\dfrac{7}{2}=\dfrac{9}{2}$

$t(-4)=3\to(-4,\ 3)\\t(4)=4\to(4,\ 4)\\\\m=\dfrac{4-3}{4-(-4)}=\dfrac{1}{8}\\\\y=\dfrac{1}{8}x+b\\\\\text{put the coordinates of the point (4,\ 4):}\\\\4=\dfrac{1}{8}(4)+b\\\\4=\dfrac{1}{2}+b\qquad\text{subtract}\ \dfrac{1}{2}\ \text{from both sides}\\\\3\dfrac{1}{2}=b\to b=3\dfrac{1}{2}=\dfrac{7}{2}\\\\t(x)=\dfrac{1}{2}x+\dfrac{7}{2}\\\\t(0)=\dfrac{1}{2}(0)+\dfrac{7}{2}=0+\dfrac{7}{2}=\dfrac{7}{2}$