Mike and Menna were instructed to graph the function y = 12 x + 1. Their graphs are shown.
The figure shows two graphs in the xy-plane. The graph on the left is labeled as Mike's Graph. The values on the x-axis range from negative 8 to 8 in increments of 2 and the values on the y-axis range from negative 8 to 8 in increments of 2. A line is shown which intersects the x-axis at negative 0.5 and y-axis at 1. The graph on the right is labeled as Menna's Graph. The values on the x-axis range from negative 8 to 8 in increments of 2 and the values on the y-axis range from negative 8 to 8 in increments of 2. A line is shown which intersects the x-axis at 2 and y-axis at 1.

Which student graphed the function correctly?
What mistake did the other student make?

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babundra210 babundra210 · Answer 1 · 2020-01-09T11:04:09+01:00

Answer:

The function is y = 2x + 1.

And mike graphed the function correctly.

Menna took the point where the function touches x-axis incorrectly.

Instead of (-1/2,0), Menna took it as (-2,0)

Step-by-step explanation:

The equation y = mx +c indicates a straight line whose slope is m

And y-intercept is c.

y-intercept is nothing but the distance between origin and point where the graph crosses the y-axis (0,c).

Now, this graph crosses x-axis when y = 0.

⇒ mx + c = 0; ⇒ x = [tex]\frac{-c}{m}[/tex].

Now, by comparing y = 2x + 1 with y = mx + c.

m=2 and c=1

⇒ the graph should crosses y-axis at (0,c) = (0,1)

And touch x-axis at ([tex]\frac{-c}{m}[/tex] , 0) = ([tex]\frac{-1}{2}[/tex],0)

⇒ mike graphed the function correctly.

Menna took the point where the function touches x-axis incorrectly.

Instead of (-1/2,0), Menna took it as (-2,0)

rshah0003 rshah0003 · Answer 2 · 2020-12-17T20:14:08+01:00

rshah0003 rshah0003

Answer:

Step-by-step explanation:

y = 2x + 1.

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