Question

A linear function has an x-intercept of 12 and a slope of StartFraction 3 Over 8 EndFraction. How does this function compare to the linear function that is represented by the table?
x
y
Negative two-thirds
Negative three-fourths
Negative one-sixth
Negative StartFraction 9 Over 16 EndFraction
One-third
Negative StartFraction 3 Over 8 EndFraction
StartFraction 5 Over 6 EndFraction
Negative StartFraction 3 Over 16 EndFraction


A It has the same slope and the same y-intercept.
B It has the same slope and a different y-intercept.
C It has the same y-intercept and a different slope.
D It has a different slope and a different y-intercept.

Answer 1

Answer:

(B) It has the same slope and a different y-intercept

Step-by-step explanation:

The table is presented below:

$\left|\begin{array}{c|c}x&y\\----&---\\-\frac{2}{3} &-\frac{3}{4}\\\\-\frac{1}{6}&-\frac{9}{16}\\\\\frac{1}{3}&-\frac{3}{8}\\\\\frac{5}{6}&-\frac{3}{16}\end{array}\right|$

Gradient:

$m=\dfrac{-\frac{9}{16}-(-\frac{3}{4})}{-\frac{1}{6}-(-\frac{2}{3})}\\=\dfrac{-\frac{9}{16}+\frac{3}{4}}{-\frac{1}{6}+\frac{2}{3}}\\=\frac{3}{16}\div \frac{1}{2}\\m=\frac{3}{8}$

Next, we determine its y-intercept

Using the pair $(-\frac{2}{3},-\frac{3}{4})}$ in y=mx+c

$-\frac{3}{4}=(\frac{3}{8})(-\frac{2}{3})+c\\-\frac{3}{4}+\frac{1}{4}=c\\c=-\frac{1}{2}$

Comparing with the linear function has an x-intercept of 12 and a slope of $\frac{3}{8}$, we find out that It has the same slope and a different y-intercept.

Option B is the correct option.

Answer 2

Answer:

B!

Step-by-step explanation: