All categories

2 years ago

T(x)=−x^+4x−3 Find two points on the graph of the parabola other than the vertex and x-intercepts.

1 answer:

6 0

The points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)

<h3>How to determine the points on the graph of the parabola other than the vertex and x-intercepts?</h3>

The equation of the parabola is given as:

t(x) = -x^2 + 4x - 3

The vertex of the parabola is the point where the graph is at the maximum or the minimum

While the x-intercept is the point where the graph crosses the x-axis i.e when y = 0

Having said that, we have the equation of the parabola to be

t(x) = -x^2 + 4x - 3

Set x = 5.

So, we have:

t(5) = -5^2 + 4 * 5 - 3

Evaluate the exponents

t(5) = -25 + 4 * 5 - 3

Evaluate the products

t(5) = -25 + 20 - 3

Evaluate the sum and the difference

t(5) = -8

Set x = 10.

So, we have:

t(10) = -10^2 + 4 * 10 - 3

Evaluate the exponents

t(10) = -100 + 4 * 10 - 3

Evaluate the products

t(10) = -100 + 40 - 3

Evaluate the sum and the difference

t(10) = -63

Hence, the points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)

You might be interested in

Find the holes<br> <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%2B3%7D%7B%28x%2B1%29%28x-2%29%7D" id="TexFormula1" t

olga55 [171]

Answer:

Step-by-step explanation:

The "holes" are the x-es where function doesn't exist.

one must not divide by 0, so given function doesn't exis if the denominator is equal 0

The product {(x+1)×(x-2)} is equal 0 if any of factors {(x+1),(x-2)} is 0, so

x + 1 = 0 or x - 2 = 0

x = - 1 or x = 2

8 0

2 years ago

The given line passes through the points (-4, -3) and (4,

drek231 [11]

Answer:

y = 2x + 11

Step-by-step explanation:

The first thing we need to do

is to find the slope of the line that passes through the points (-4,-3) and (4,1)

Mathematically, that would be;

m = y2-y1/(x2-x1)

where (x1,y1) = (-4,-3) and (x2,y2) = (4,1)

substituting these. values we have;

m = (1-(-3))/(4-(-4)) = 4/8 = 1/2 or 0.5

Now we are told this line is perpendicular to another line that passes through another point.

We can find the slope of this other line

Since both lines are perpendicular, the product of their slope is -1.

Thus , -0.5 * m = -1

m = -1/-0.5 = 2

So the slope of the other line is 2

Using the point-slope form;

y-y1= m(x-x1)

The point for the other line is (-4,3)

So the equation will be

y-3 = 2(x+4)

y-3 = 2x + 8

y = 2x + 11

8 0

3 years ago

Adam currently runs about 40 miles per week, and he wants to increase his weekly mileage by 10%. How many miles will Adam run pe

nadya68 [22]

it would be 50miles i think or it would be 44

8 0

2 years ago

An investment broker puts 1/12 of his paycheck into a retirement account every quarter? How much of his $84,000 salary is put in

posledela

Answer:

<u>$7000</u> of his salary is put into this account each quarter.

Step-by-step explanation:

Given:

An investment broker puts 1/12 of his paycheck into a retirement account every quarter.

His salary is $84,000.

Now, to find how much is put into this account each quarter.

As given salary = $84,000.

So, amount to put into this account every quarter =

$\frac{1}{12} of\ \$84,000.$

$=\frac{1}{12} \times 84000$

$=\frac{84000}{12}$

$=\$7000.$

Therefore, $7000 of his salary is put into this account each quarter.

6 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago