Help me fast times-a-ticking

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

8 0

No, because it has a constant rate of change

By looking at this table

The y value changes at a constant multiple of 2

6 4

7 2 (4-2 is a difference of 2)

8 0 (2-0 is a difference of 2)

9 -2 (0-2 is a difference of 2)

This means it is Linear

mash [69]3 years ago

4 0

No, because it has a constant rate of charge.

Rate of change of function: $\dfrac{y_2-y_1}{x_2-x_1}$

$\begin{array}{ccc}6\to4\\7\to2\end{array}\Rightarrow\dfrac{2-4}{7-6}=\dfrac{-2}{1}=-2\\\\\begin{array}{ccc}7\to2\\8\to0\end{array}\Rightarrow\dfrac{0-2}{8-7}=\dfrac{-2}{1}=-2\\\\\begin{array}{ccc}8\to0\\9\to-2\end{array}\Rightarrow\dfrac{-2-0}{9-8}=\dfrac{-2}{1}=-2\\\\$

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Write the equation of a parabola having the vertex (1, −2) and containing the point (3, 6) in vertex form. Then, rewrite the equ

In-s [12.5K]

PART A

The equation of the parabola in vertex form is given by the formula,

$y - k = a {(x - h)}^{2}$

where

$(h,k)=(1,-2)$

is the vertex of the parabola.

We substitute these values to obtain,

$y + 2 = a {(x - 1)}^{2}$

The point, (3,6) lies on the parabola.

It must therefore satisfy its equation.

$6 + 2 = a {(3 - 1)}^{2}$

$8= a {(2)}^{2}$

$8=4a$

$a = 2$
Hence the equation of the parabola in vertex form is

$y + 2 = 2 {(x - 1)}^{2}$

PART B

To obtain the equation of the parabola in standard form, we expand the vertex form of the equation.

$y + 2 = 2{(x - 1)}^{2}$

This implies that

$y + 2 = 2(x - 1)(x - 1)$

We expand to obtain,

$y + 2 = 2( {x}^{2} - 2x + 1)$

This will give us,

$y + 2 = 2 {x}^{2} - 4x + 2$

$y = {x}^{2} - 4x$

This equation is now in the form,

$y = a {x}^{2} + bx + c$
where

$a=1,b=-4,c=0$

This is the standard form

7 0

2 years ago

Which equation describes a line parallel<br> to y= 4x+ 5 through point (-2, 1)?

Tema [17]

Hi there!

For a line to be PARALLEL, it must contain the same slope.

The given line has a slope of 4, so we can use the point-slope formula to solve:

$\large\boxed{y - y_1 = m(x - x_1)}$