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3 years ago

The line with gradient-4 which cuts the y axis at -2

Mathematics

1 answer:

Helen [10]3 years ago

8 0

Answer:

• Considering the general equation of a line

$y = mx + b$

m is the gradient and b is the y-intercept.

Substitute:

${ \underline{ \underline{ \: \: y = - 4x - 2 \: \: }}}$

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Find an equation of the line described below. Write the equation in slope-intercept form (solved for y), when possible.

Vinil7 [7]

Answer:

Step-by-step explanation:

<u>Equation of a line</u>

A line can be completely defined by two points. Suppose we know the line passes through points A(x1,y1) and B(x2,y2).

The equation for a line can be written as:

$y=mx+b$

Where m is the slope and b is the y-intercept. Both values can be determined by using the coordinates of the given points.

First, determine the slope with the equation:

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

The points are: A(15,3) B(3,15)

$\displaystyle m=\frac{15-3}{3-15}=\frac{12}{-12}=-1$

The equation of the line can be written as:

$y=-x+b$

Now, use any point to determine the value of b. Substitute (15,3):

$3=-15+b$

Solve for b:

b=18

The equation of the line is

y=-x+18

The slope is -1 and the y-intercept is 18.

3 0

3 years ago

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Natasha2012 [34]

Answer:

$x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$ are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

$D=\sqrt{(-10)^2-4(2)(-3)}$

$\Rightarrow D=\sqrt{100+24}=\sqrt{124}$

Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

$x=\frac{50\pm\sqrt{31}}{2}$

Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

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4 years ago

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4 years ago

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4 years ago

The line with gradient-4 which cuts the y axis at -2​

The line with gradient-4 which cuts the y axis at -2