Question

Write a linear equation that intersects ytte points. Then willes second linear equation that intersecta y at just one point, and ther equation that does not intersecty Expten how you found the linear equations 40

Answer 1

Answer:

$(a)\ y = 2x +8$

$(b)\ y = 0$

$(c)\ y = -5$

Step-by-step explanation:

Given

$y = x^2$

First, graph $y = x^2$ (See attachment)

We are not limited to any particular linear equation; as long as we meet the required condition

Solving (a): Equation of a line that intersects $y = x^2$ at two points

Select any two points on opposite sides of the curve

The selected points are:

$(x_1,y_1) = (-2,4)$

$(x_2,y_2) = (4,16)$

Calculate slope (m)

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{16 - 4}{4 - -2}$

$m = \frac{12}{6}$

$m = 2$

The equation is then calculated using:

$y = m(x - x_1) + y_1$

This gives:

$y = 2(x - -2) + 4$

$y = 2(x +2) + 4$

Open bracket

$y = 2x +4 + 4$

$y = 2x +8$

Solving (b): Equation of a line that intersects $y = x^2$ at one point

A good illustration of this is x-axis

Because it intersects with the line at just one point, which is origin (0,0).

The equation of line at this point is:

$y = 0$

Solving (c): Equation of a line that does not intersect $y = x^2$

Here, we make use of any horizontal line below the x-axis

Select any two horizontal points below the x-axis

$(x_1,y_1) = (-5,-5)$

$(x_2,y_2) = (5,-5)$

Calculate slope

$m = \frac{-5--5}{5--5}$

$m = \frac{0}{10}$

$m = 0$

The equation is then calculated using:

$y = m(x - x_1) + y_1$

$y = 0(x - -5) + (-5)$

$y = 0-5$

$y = -5$