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2 years ago

What is the equation of the graph

Mathematics

1 answer:

AveGali [126]2 years ago

5 0

Answer:

$y = 2x^2 + 1$

Step-by-step explanation:

The graph you see there is called a parabola. The general equation for the graph is as below

$y = a*x^2 + b$

To find the equation we need to find the constants a and b. The constant b is just how much we're lifting the parabola by. Notice it's lifted by 1 on the y axis.

To find a it's a little more tricky. Let's use the graph to find a value for a by plugging in values we know. We know that b is 1 from the previous step, and we know that when x=1, y=3. Let's use that!

$3 = a * (1)^2 + 1\\2 = a$

Awesome, we've found both values. And we can write the result.

$y = 2x^2 + 1$

I'll include a plotted graph with our equation just so you can verify it is indeed the same.

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7. What is the cardinality of each of the following sets?

zysi [14]

Answer:

The cardinality of set (a) is 0,

the cardinality of set (b) is 1

and

the cardinality of set (c) is 3.

Step-by-step explanation: We are given to find the cardinality of each of the following sets :

(a) { }.

(b) { { } }.

We know that

CARDINALITY of a set is the number of elements present in the set.

(a) The given set is A = { }.

The set A is an empty set, so it does not contain any element. Hence, the cardinality of the set A is 0.

(b) The given set is B = { { } }.

The set is B is singleton set, contains only one element (that is empty set). So, the cardinality of the set B is 1.

(c) The given set is C = {a, {a}, {a, {a}} }.

The set C has three elements, a, the set {a} and the set {a, {a}}. So, the the cardinality of set C is 3.

Thus,

the cardinality of set (a) is 0,

the cardinality of set (b) is 1

and

the cardinality of set (c) is 3.

8 0

2 years ago

5. (9) Letf- ((-2, 3), (-1, 1), (0, 0), (1,-1), (2,-3)) and let g- ((-3, 1), (-1, -2), (0, 2), (2, 2), (3, 1)j. Find: a. (g f (0

pashok25 [27]

Answer: g(f(0)) = 2 and (f ° g)(2) = -3.

Step-by-step explanation: We are given the following two functions in the form of ordered pairs :

f = {(-2, 3), (-1, 1), (0, 0), (1,-1), (2,-3)}

g = {(-3, 1), (-1, -2), (0, 2), (2, 2), (3, 1)} .

We are to find g(f(0)) and (f ° g)(2).

We know that, for any two functions p(x) and q(x), the composition of functions is defined as

$(p\circ q)(x)=p(q(x)).$

From the given information, we note that

f(0) = 0, g(0) = 2, g(2) = 2 and f(2) = -3.

So, we get

$g(f(0))=g(0)=2,\\\\(f\circ g)(2)=f(g(2))=f(2)=-3.$

Thus, g(f(0)) = 2 and (f ° g)(2) = -3.

8 0

3 years ago

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
$

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
$\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2$

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²

$c^2+4c+4=0
\\(c+2)^2=0
\\c=-2$

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$