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1 year ago

Choose all sequences of transformations that produce the same image of a given figure.

1 answer:

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<em><u>Your answer: </u></em> a reflection across the y-axis followed by a clockwise rotation 90° about the origin. A clockwise rotation 90° about the origin followed by a reflection across the x-axis. A counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis. A reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin.

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(i)Express 2x² – 4x + 1 in the form a(x+ b)² + c and hence state the coordinates of the minimum point, A, on the curve y= 2x² 4x

earnstyle [38]

Answer:

(i). $y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1$ . Point $A$ is at $(1, \, -1)$ .

(ii). Point $Q$ is at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ .

(iii). $\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ (slope-intercept form) or equivalently $x + 5\, y - 17 = 0$ (standard form.)

Step-by-step explanation:

<h3>Coordinates of the Extrema</h3>

Note, that when $a(x + b)^2 + c$ is expanded, the expression would become $a\, x^2 + 2\, a\, b\, x + a\, b^2 + c$ .

Compare this expression to the original $2\, x^2 - 4\, x + 1$ . In particular, try to match the coefficients of the $x^2$ terms and the $x$ terms, as well as the constant terms.

For the $x^2$ coefficients: $a = 2$ .
For the $x$ coefficients: $2\, a\, b = - 4$ . Since $a = 2$ , solving for $b$ gives $b = -1$ .
For the constant terms: $a \, b^2 + c = 1$ . Since $a = 2$ and $b = -1$ , solving for $c$ gives $c =-1$ .

Hence, the original expression for the parabola is equivalent to $y = 2\, (x - 1)^2 - 1$ .

For a parabola in the vertex form $y = a\, (x + b)^2 + c$ , the vertex (which, depending on $a$ , can either be a minimum or a maximum,) would be $(-b,\, c)$ . For this parabola, that point would be $(1,\, -1)$ .

<h3>Coordinates of the Two Intersections</h3>

Assume $(m,\, n)$ is an intersection of the graphs of the two functions $y = 2\, x^2- 4\, x + 1$ and $x -y + 4 = 0$ . Setting $x$ to $m$ , and $y$ to $n$ should make sure that both equations still hold. That is:

$\displaystyle \left\lbrace \begin{aligned}& n = 2\, m^2 - 4\, m + 1 \\ & m - n + 4 = 0\end{aligned}\right.$ .

Take the sum of these two equations to eliminate the variable $n$ :

$n + (m - n + 4) = 2\, m^2 - 4\, m + 1$ .

Simplify and solve for $m$ :

$2\, m^2 - 5\, m -3 = 0$ .

$(2\, m + 1)\, (m - 3) = 0$ .

There are two possible solutions: $m = -1/2$ and $m = 3$ . For each possible $m$ , substitute back to either of the two equations to find the value of $n$ .

$\displaystyle m = -\frac{1}{2}$ corresponds to $n = \displaystyle \frac{7}{2}$ .
$m = 3$ corresponds to $n = 7$ .

Hence, the two intersections are at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ and $(3,\, 7)$ , respectively.

<h3>Line Joining Point Q and the Midpoint of Segment AP</h3>

The coordinates of point $A$ and point $P$ each have two components.

For point $A$ , the $x$ -component is $1$ while the $y$ -component is $(-1)$ .
For point $P$ , the $x$ -component is $3$ while the $y$ -component is $7$ .

Let $M$ denote the midpoint of segment $AP$ . The $x$ -component of point $M$ would be $(1 + 3) / 2 = 2$ , the average of the $x$ -components of point $A$ and point $P$ .

Similarly, the $y$ -component of point $M$ would be $((-1) + 7) / 2 = 3$ , the average of the $y\!$ -components of point $A$ and point $P$ .

Hence, the midpoint of segment $AP$ would be at $(2,\, 3)$ .

The slope of the line joining $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ (the coordinates of point $Q$ ) and $(2,\, 3)$ (the midpoint of segment $AP$ ) would be:

$\displaystyle \frac{\text{Change in $y$}}{\text{Change in $x$}} = \frac{3 - (7/2)}{2 - (-1/2)} = \frac{1}{5}$ .

Point $(2,\, 3)$ (the midpoint of segment $AP$ ) is a point on that line. The point-slope form of this line would be:

$\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)$ .

Rearrange to obtain the slope-intercept form, as well as the standard form of this line:

$\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ .

$x + 5\, y - 17 = 0$ .

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1 year ago

which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

givi [52]

Which expression is equivalent to 100 n2 − 1? (10n)2 − (1)2 (10n2)2 − (1)2 (50n)2 − (1)2 (50n2)2 − (1)2

Regroup:
100n^2-1

=10^2n^2-1
=(10n)^2-1^2

So the correct answer is :
(10n)^2-1^2

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