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Fed [463]
3 years ago
12

Graph the linear equation. Find three points that solve the equation, then plot on the graph.

Mathematics
1 answer:
Illusion [34]3 years ago
8 0

Answers:

Three points that solve the equation: (1, \frac{19}{4} ) , (2, 5), (3, \frac{21}{4} )

The graph is shown in the attached pictures.

NOTE: The first picture is the graph of the equation along with the plotted points, and the second one shows the work for those three points.

Step-by-step explanation:

1. To graph this equation, an easier way to do it would be to convert to slope-intercept form so we can graph knowing the y-intercept and the slope. Do this by isolating the y on the left side like so:

x-4y=-18\\-4y = -x -18\\y = \frac{-x}{-4} -\frac{18}{-4} \\y = \frac{1}{4} x + \frac{9}{2}

Remember that slope-intercept form is in y = mx + b format, and that m is the slope and b is the y-intercept. With this information, we know that (0, \frac{9}{2}) is the y-intercept and \frac{1}{4} is the slope of this equation. We can plot the point (0, \frac{9}{2}) on the graph, and then use the slope of \frac{1}{4} from there to graph other points and form a line. (When I graphed the line, I didn't include these "other points" so it wasn't confusing to locate which points were the three solutions listed.)

2. Points that solve an equation - or solutions - are also points that the line of the equation intersects. So, what we can do is form a table, plug in some x values into the equation, and solve for a y-value. The x and y values will form a point that is on the graph, thus they are solutions. (Please look at the second picture for work and clarification.) After identifying these points, just plot them on the graph and label them (as shown in the first picture).

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Read 2 more answers
An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if p
horrorfan [7]

Answer:

a = -0.3575

Step-by-step explanation:

The points A and D lie on the x-axis, this means that they are the x-intercepts of the parabola, and therefore we can find their location.

The points A and B are located where

y=a(x+1)(x-5)=0

This gives

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y=5

Now given the coordinates of A, we are in position to find the coordinates of the point B. Point B must have y coordinate of y=2 (because the base of the trapezoid is at y=0), and the x coordinate of B, looking at the figure, must be x coordinate of A plus horizontal distance between A and B, i.e

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Thus the coordinates of B are:

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Now this point B lies on the parabola, and therefore it must satisfy the equation  y=a(x+1)(x-5).

Thus

2=a((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)

Therefore

a=\frac{2}{((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)}

\boxed{a=-0.3575}

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