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tia_tia [17]
3 years ago
5

In what ways does adding a constant or coefficient to a quadratic equation affect the graph?

Mathematics
2 answers:
nekit [7.7K]3 years ago
4 0
When you add a constant to a quadratic equation, the vertex of the graph will change depending on whether the constant is positive or negative.
When you add a coefficient to a quadratic equation, the graph will widen or narrow depending on whether the coefficient is positive or negative.
The axis of symmetry always passes through the vertex.
The key features of a quadratic graph are vertex, axis of symmetry, intercepts, focus, latus rectum, and directrix.
Radda [10]3 years ago
4 0
<h2>Answer with explanation:</h2>

<u>Ques 1)</u>

The quadratic equation is given by:

y=a(x-h)^2+k

                      Vertex is: (h,k)

Hence, on adding a simple constant the y-coordinate of the vertex gets changed , since there is a change in the term "k"

and adding a coefficient to the quadratic equation will change the graph in the manner that it could be wide or narrow depending on whether the coefficient is greater than 1 or less than 1 and also the sign of the coefficient describes whether the graph is open upward or downward.

<u>Ques 2)</u>

The axis of symmetry and the vertex are related in the manner that the axis of symmetry always passes through the vertex of the graph.

i.e. if the graph is a upward or downward open parabola then axis of symmetry is: y=k

and if the graph is a right or left open parabola then axis of symmetry is: x=h

where the vertex is: (h,k)

<u>Ques 3)</u>

The key features of a quadratic graph are:

1)   Vertex.

2)  Axis of symmetry.

3) x-intercept( also known as zeros)

4)  y-intercept.

5)  Focus

6)  Directrix.

7)  Latus Rectum.

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Solve the system of eqautions <br> y=x2-5x+7 , y=2x+1
IgorC [24]

Answer:

Step-by-step explanation:

The system of equations is given as

y=x^2-5x+7 - - - - - - - - - - 1

y=2x+1 - - - - - - - - - - - - - - 2

We would equate equation 1 to equation 2, it becomes

x^2-5x+7 = 2x+1

x^2-5x+7- 2x - 1 = 0

x^2-5x+7- 2x - 1 = 0

x^2 - 5x - 2x + 7 - 1 = 0

x^2 - 7x + 6 = 0

We would use the factorization method of solving quadratic equations.

x^2 - 6x - x + 6 = 0

x(x - 6) - 1(x - 6)

(x - 6)(x - 1) = 0

x - 6 = 0 or x - 1 = 0

x = 6 or x = 1

Substituting both values of x into equation 2, it becomes

For x = 6,

y=2×6 + 1 = 12 + 1 = 13

y = 13

For x = 1,

y=2 × 1 +1 = 2 + 1 = 3

y = 3

3 0
2 years ago
Use a table of values with at least 5 values to graph the following function:
34kurt

The given expression :

y=(\frac{1}{2})^x

For coordinates:

put x = 0 then :

\begin{gathered} y=(\frac{1}{2})^0 \\ y=1 \end{gathered}

Coordinate : (x, y) = (0, 1)

Put x= 1 and simplify :

\begin{gathered} y=(\frac{1}{2})^1 \\ y=\frac{1}{2} \\ y=0.5 \end{gathered}

Coordinate : (x, y) = ( 1, 0.5)

Put x = (-2) and simplify :

\begin{gathered} y=(\frac{1}{2})^{-2} \\ y=\frac{1^{-2}}{2^{-2}} \\ y=\frac{2^2}{1^2} \\ y=2^2 \\ y=4 \end{gathered}

Coordinate : (x, y) = ( -2, 4)

Put x = (-3) and simplify :

\begin{gathered} y=(\frac{1}{2})^{-3} \\ y=\frac{1^{-3}}{2^{-3}} \\ y=\frac{2^3}{1^3} \\ y=2^3 \\ y=8 \end{gathered}

Coordinate : (x, y) = (-3, 8)

Substitute x = (-1) and simplify :

\begin{gathered} y=(\frac{1}{2})^x \\ y=(\frac{1}{2})^{-1} \\ y=\frac{1^{-1}}{2^{-1}} \\ y=\frac{2}{1} \\ y=2 \end{gathered}

Coordinate : (x, y) = ( -1, 2)

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The graph is :

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Find the area.. of the figure below...
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Read 2 more answers
When electricity (the flow of electrons) is passed through a solution, it causes an oxidation-reduction (redox) reaction to occu
oee [108]

Answer:

a. 135 g

b. 60.6 min

Step-by-step explanation:

a. What mass of Cu(s) is electroplated by running 28.5 A of current through a Cu2+ (aq) solution for 4.00 h? Express your answer to three significant figures and include the appropriate units.

The chemical equation for the reaction is given below

Cu²⁺(aq) + 2e⁻ → Cu(s)

We find the number of moles of Cu that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 28.5 A and t = time = 4.00 h = 4.00 × 60 min/h × 60 s/min = ‭14,400‬ s

So, n = It/F = 28.5 A × ‭14,400‬ s/96485 C/mol = ‭410,400‬ C/96485 C/mol = 4.254 mol

Since 2 moles of electrons deposits 1 mol of Cu, then 4.254 mol of electrons deposits 4.254 mol × 1 mol of Cu/2 mol = 2.127 mol of Cu

Now, number of moles of Cu = n' = m/M where m = mass of copper and M = molar mass of Cu = 63.546 g/mol

So, m = n'M

= 2.127 mol × 63.546 g/mol

= 135.15 g

≅ 135 g to 3 significant figures

b. How many minutes will it take to electroplate 37.1 g of gold by running 5.00 A of current through a solution of Au+(aq)?

The chemical equation for the reaction is given below

Au⁺(aq) + e⁻ → Au(s)

We need to find the number of moles of Au in 37.1 g

So, number of moles of Au = n = m/M where m = mass of gold = 37.1 g and M = molar mass of Au = 196.97 g/mol

So, n = m/M = 37.1 g/196.97 g/mol = 0.188 mol

Since 1 mol of Au is deposited  by 1 moles of electrons, then 0.188 mol of Au deposits 0.188 mol of Au × 1 mol of electrons/1 mol of Au = 0.188 mol of electrons

We find the time it takes to deposit 0.188 mol of electrons that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 5.00 A and t = time

So, t = nF/It

= 0.188 mol × 96485 C/mol ÷ 5.00 A

= ‭18173.30‬ C/5.00 A

= 3634.66 s

= 3634.66 s × 1min/60 s

= 60.58 min

≅ 60.6 min to 3 significant figures

6 0
3 years ago
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