Ask question

Ask question

All categories

3 years ago

5

Find the quadratic function y=f(x) whose graph has a vertex (−3,3) and passes through the point (−6,0). What is the functi

on in standard form?

1 answer:

adell [148]3 years ago

4 0

Answer:

f(x) = -1/3x² -2x

Step-by-step explanation:

For vertex (h, k) and scale factor "a", the vertex form of a quadratic is ...

y = a(x -h)² +k

For the given vertex, the vertex form function with scale factor "a" is ...

f(x) = a(x +3)² +3

If that goes through the point (-6, 0), then we have ...

f(-6) = a(-6 +3)² +3 = 0

9a = -3 . . . . . . . subtract 3

a = -1/3 . . . . . . . divide by 9

In standard form, the equation is ...

f(x) = (-1/3)x² -2x

You might be interested in

Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

<h3>1.</h3>

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

<h3>2.</h3>

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

<h3>3.</h3>

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.

$\angle 9=180-59-59=62$ , by interior angles theorem.

$\angle 8 =62$ , by vertical angles theorem.

$\angle 10 = 180- \angle 9=180-62=118\°$ , by supplementary angles.

$\angle 7 = 118\°$ , by vertical angles theorem.

$\angle 5=90-59=31$ , by complementary angles.

$\angle 2 = \angle 5 = 31\°$ , by alternate interior angles.

$\angle 1 = 59\°$ , by complementary angles.

<h3>4.</h3>

$m\angle BCD=90\°$ , because it's one of the four interior angles of a rectangle, which by deifnition are equal to 90°.

$m\angle ABD = 6\° = m\angle BDC$ , by alternate interior angles and by given. $m\angle CBE=90-6=84$ , by complementary angles.

$m\angle ADE=90-6=84$ , by complementary angles.

$m\angle AEB=180-6-6=168\°$ , by interior angles theorem.

$m\angle DEA=180-168=12$ , by supplementary angles.

<h3>5.</h3>

$m\angle JMK=180-126=54$ , by supplementary angles.

$m\angle JKH=\frac{180-54}{2}=\frac{126}{2}=63$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle HLK=90\°$ , by definition of rectangle.

$m\angle HJL=\frac{180-126}{2}=27$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle LHK=90-27=63$ , by complementary angles.

$m\angle = JLK= m\angle HJL=27$ , by alternate interior angles.

<h3>6.</h3>

The figure is a rectangle, which means its opposite sides are equal, so

$WZ=XY\\7x-6=3x+14\\7x-3x=14+6\\4x=20\\x=\frac{20}{4}\\ x=5$

Then, we replace this value in the expression of side WZ

$WZ=7x-6=7(5)-6=35-6=29$

Therefore, side WZ is 29 units long.

<h3>7.</h3>

We know that the diagonals of a rectangle are congruent, so

$SQ=PR\\11x-26=5x+28\\11x-5x=28+26\\6x=54\\x=\frac{54}{6}\\ x=9$

Then,

$PR=5x+28=5(9)+28=45+28=73$

Therefore, side PR is 73 units long.

3 0

3 years ago

Can u please help me with question c

densk [106]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the y-axis

a point (x, y ) → (- x, y )

Hence

A'(-5, - 3 ) → A''(5, - 3 )

B'(4, 4 ) → B''(- 4, 4 )

C'(4, 2 ) → C''(- 4, 2 )

D'(- 2, - 1 ) → D''(2, - 1 )

8 0

3 years ago

Can someone help me with this question if two figures are similar but not congruent, how are they alike, and how are they differ

kenny6666 [7]

Two figures are similar if they have the same shape but not necessarily the same size although they are the same shape they are different sizes

3 0

2 years ago

find the number 15 balls consisting of 5 black, 5 blue nd 5 white balls from a box which has 6 black, 7 blue and 8 white balls (

AleksAgata [21]

Answer:

Step-by-step explanation:

A bag contains 4 blue, 5 white, and 6 green balls. ... There are 5 red and 15 black balls in a box, two are picked up at random

4 0

3 years ago

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

7 0

3 years ago