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3 years ago

Identify the system of equations given in the graph below.

Mathematics

1 answer:

Pavel [41]3 years ago

7 0

Hi there!

$\large\boxed{\left \{ {{y=3x + 1} \atop {y = 3x - 1}} \right}$

From the graph, we can see that both lines are parallel. We can determine the slope of one line to find the slope of the other.

Use the slope formula to determine the slope of a line. In this case, I am using the blue line as a reference:

$\text{Slope } = \frac{ y_2 - y_1 } { x_2 - x_1 }$

$m = \frac{ 3 - 1 } { 1 - 0 } = 3$

The slope of both lines is 3.

Let's find the equation of the blue line by finding its y-intercept to solve for the "b" value in the equation y= mx + b:

y-intercept is when x = 0, or at (0, 1). Therefore:

Blue line equation: y = 3x + 1

The red line intersects the y-axis at (0, -1), therefore:

Red line equation: y = 3x - 1

The systems of equations is:

y = 3x + 1

y = 3x - 1

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Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Blizzard [7]

Answer:

One possible equation is $f(x) = (x + 3)\, (x - 5)$ , which is equivalent to $f(x) = x^{2} - 2\, x - 15$ .

Step-by-step explanation:

The factor theorem states that if $x = x_{0}$ (where $x_{0}$ is a constant) is a root of a function, $(x - x_{0})$ would be a factor of that function.

The question states that $(-3,\, 0)$ and $(5,\, 0)$ are $x$ -intercepts of this function. In other words, $x = -3$ and $x = 5$ would both set the value of this quadratic function to $0$ . Thus, $x = -3\!$ and $x = 5\!$ would be two roots of this function.

By the factor theorem, $(x - (-3))$ and $(x - 5)$ would be two factors of this function.

Because the function in this question is quadratic, $(x - (-3))$ and $(x - 5)$ would be the only two factors of this function. In other words, for some constant $a$ ( $a \ne 0$ ):

$f(x) = a\, (x - (-3))\, (x - 5)$ .

Simplify to obtain:

$f(x) = a\, (x + 3)\, (x - 5)$ .

Expand this expression to obtain:

$f(x) = a\, (x^{2} - 2\, x - 15)$ .

(Quadratic functions are polynomials of degree two. If this function has any factor other than $(x - (-3))$ and $(x - 5)$ , expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of $a$ corresponds to a distinct quadratic function with $x$ -intercepts $(-3,\, 0)$ and $(5,\, 0)$ . For example, with $a = 1$ :

$f(x) = (x + 3)\, (x - 5)$ , or equivalently,

$f(x) = x^{2} - 2\, x - 15$ .

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