How do you graph 2x-y=3

Mathematics

2 answers:

Aneli [31]2 years ago

6 0

Basically you have to put the equation into slope intercept form which first you move 2x over making it a negative so you have
-y= -2x+3. Then you can’t have a negative y so divide it by -1 on both sides making the final equation y=2x-3. Since slope intercept form is y=mx+b you know the slope (m) is 2 and the y-intercept (b) is -3. Then you just make the coordinate (0,-3) for the y-intercept and from there create a line that has a slope of 2 in which it goes through the point you’ve made. Hope this helps.

Alexxandr [17]2 years ago

3 0

To make this equation easier to graph, convert it to slope-intercept form.
Slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

To get 2x - y = 3 into slope-intercept form, isolate y.

2x - y = 3
-y = 3 - 2x
y = -3 + 2x
y = 2x - 3

As you can see, this graph has a slope of 2 and a y-intercept of -3.

The first step to graphing is plotting the y-intercept, which is (0, -3).

From there, figure out the slope. It's positive 2, so the graph will slant upwards. That means you can count 2 units up and 1 unit over and plot that point. Keep counting 2 units up and 1 over to begin forming the graph.

Finally, connect the coordinates, and you've graphed 2x - y = 3.

I've attached the graph below.

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Find the range of p in equation p(x+1) (x-3)=x-4p-2 has no real roots.

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Answer:

$\displaystyle p > \frac{1}{4}$ .

Step-by-step explanation:

Expand the left-hand side of this equation:

$p\, (x + 1)\, (x - 3) = p\, \left(x^2 - 2\, x - 3\right) = p\, x^2 - 2\, p\, x - 3\, p$ .

Collect the terms, so that this quadratic equation is in the form $a\, x^2 + b\, x+ c = 0$ :

$p\, x^2 - 2\, p\, x - 3\, p = x - 4\, p - 2$ .

$p\, x^2 - (2\, p + 1)\, x + (p + 2) = 0$ .

In this equation:

$a = p$ .
$b = -(2\, p + 1)$ .
$c = p + 2$ .

Calculate the quadratic discriminant of this quadratic equation:

$\begin{aligned}b^2 - 4\, a\, c &= (-(2\, p + 1))^2 - 4\, p\, (p + 2) \\ &= 4\, p^2 + 4\, p + 1 - 4\, p^2- 8\, p = -4\, p + 1\end{aligned}$ .

A quadratic equation has no real root if its quadratic discriminant is less then zero. As a result, this quadratic equation will have no real root when $-4\, p + 1 < 0$ . Solve for the range of $p$ :