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igor_vitrenko [27]
3 years ago
12

Find the quadratic function y=​f(x) whose graph has a vertex ​(−3​,4​) and passes through the point ​(−7​,0). Write the function

in standard form.
Mathematics
1 answer:
olganol [36]3 years ago
5 0

Answer:

Step-by-step explanation:

This is a parabola since a quadratic is a parabola.  The standard form for a parabola is y = ax² + bx + c

but before we do that, we will use the vertex form, since it will make our work easier at the beginning.  

First and foremost, when we plot the vertex and the given point, the vertex is higher up than is the point; that means that this parabola opens upside down, and its vertex form will be

y=-|a|(x - h)² + k

The absolute value is out in front of the a, so we know that the value of a is positive, but the quadratic itself is negative (upside down) and we will find that math takes care of that negative that needs to be out front.  So we need to solve for a by filling in the x, y, h, and k values from the point and the vertex:  x = -7, y = 0, h = -3, k = 4

0 = a(-7 - (-3))² + 4 and

0 = a(-7 + 3)² + 4 and

0 = a(-4)² + 4 and

0 = a(16) + 4 and

0 = 16a + 4 and

-4 = 16a so

a=-\frac{1}{4}

Now that we know a, we can plug it back into the vertex form and then put it into standard form from there.

y=-\frac{1}{4}(x+3)^2+4

Now we will FOIL out what's inside the parenthesis to get

y=-\frac{1}{4}(x^2+6x+9)+4

Simplify by distributing the -1/4 into the parenthesis:

y=-\frac{1}{4}x^2-\frac{3}{2}x-\frac{9}{4}+4

Combine like terms to get

y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{7}{4}

And there you go!

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Hello everyone! Can you please help me with this problem, and when available, will give out brainliest! Thnx!
Pachacha [2.7K]

Answer:

The first first box needs  600 cm^2 more brown paper for wrapping

Step-by-step explanation:

Given:

First box:

Square base

Length of the base side =  30 cm

Height of the pyramid = 45 cm

Second Box:

octagonal base

base perimeter =  120 cm

width = 10 cm

Height =  35 cm

To Find:

Which box requires more paper to wrap, and by how much?

Solution:

To Find the amount are brown paper needed to wrap, lets find the surface area of the pyramids

Step: Finding the surface area of the box 1

The First box is square pyramid:

= base square area + 4 x area of side triangle

Base square area:

Area of the square  = side x side

=  30 x 30

=900 centimetre square

Area of triangle

=\frac{1}{2} base \times height

=>\frac{1}{2} (30 \times 45)

=>\frac{1350}{2}

=>675 centimetre square

Thus the total brown paper needed for wrapping the first box

= 900 + 675

= 1575  centimetre square--------------------------------(1)

Step 2: Finding the surface area of the box 2

The second box is octagonal pyramid:

= base octagonal area + 8 x area of side triangle

Base octagonal area:

Area of the base of the octagon = 8 x area of one triangle

area of one triangle =  \frac{1}{2} base \times  height

= \frac{1}{2} 15 \times 10

= \frac{150}{2}

=75 centimetre square

Area of the base of the octagon = 8 x 75 = 600 centimetre square

Area of the side triangles  

=   \frac{1}{2} base \times  height

= \frac{1}{2} (45 \times 15)

=\frac{675}{2}

= 337.5 centimetre square

Thus the total surface of the octagonal pyramid is  

= 600 +375  

=975 centimetres square -------------------------------(2)

On comparing (1) and(2)

1575 -975 = 600

The first box requires 600 cm^2 more brown paper for wrapping

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