Below are two different functions, f(x) and g(x). what can be determined about their y-intercepts?

f(x) = 4x − 1

chart:
x g(x)
1 3
4 6
7 9

answer choices:
a. the function f(x) has a higher y-intercept.
b. the function g(x) has a higher y-intercept.
c. they both have the same y-intercept.
d. the relationship between y-intercepts cannot be determined

It can be determined about their y-intercepts that b. The function g(x) has a higher y-intercept.

Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates

General formula

or

y = mx + c

Where

m = straight-line gradient which is the slope of the line

x1, y1 = the Cartesian coordinate that is crossed by the line

c = constant

The formula for a gradient (m) between 2 points in a line

m = Δy / Δx

To find the equation of the line which has the point of intersection with the highest line of y then we enter the value of x in the equation = 0, so we know which value of y is greater

In the problem there is one known line equation (f (x)) and one line equation known for the value of x and its function (g (x))

For equation f (x) the value of the intersection with the y axis is: (x = 0)

f (x) = 4x - 1 (f (x) = y)

y = 4x - 1

y = 4.0 - 1

y = -1

For equation g (x), we first look for the equation by finding the gradient first

we specify the points: (4, 6) and (7, 9)

then the price gradient (m)

m = 3/3

m = 1

We enter in the general equation with point (4,6)

y-6 = 1 (x-4)

y-6 = x-4

y = x + 2

y = g (x) = x + 2

We enter the value x = 0 to get the value of the intersection with the y axis:

y = x + 2

y = 2

These two results show that: b. The function g (x) has a higher y-intercept.

F (x) = x2 + 1 g (x) = 5 - x

htps: //link

the inverse of the function f (x) = 2x-10

link

domain of the function

link

Keywords: function, y-intercept.

good luck on your studies! : )

david edward.

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: