LOTS OF POINTS!?!?!?Write a linear function f with the values f(5)=−1 and f(0)=−1.
f(x)=

Answer:

The euation of the linear function is:

Step-by-step explanation:

As the function has the values

Thus, the two points are:

Finding the slope between (5, -1) and (0, -1)

[tex]\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\left(x_1,\:y_1\right)=\left(5,\:-1\right),\:\left(x_2,\:y_2\right)=\left(0,\:-1\right)[/tex]

[tex]m=\frac{-1-\left(-1\right)}{0-5}[/tex]

[tex]m=0[/tex]

We know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept

We also know that the y-intercept can be computed by setting x=0 and finding the corresponding value of y.

so at x = 0, y = -1

Thus, the y-intercept 'b' = -1

Now, substituting m = 0 and b = -1 in the slope-intercept form of the line equation

y = mx+b

y = 0x + -1

y = -1

Therefore, the euation of the linear function is:

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Captionless Image

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**PLEASE HELP**
Find the equation, f(x)=a(x-h)^2+k, for a parabola that passes through the point (0,0) and has a (-3,-6) as its vertex. What is the standard form of the equation?

A.) The vertex form of the equation is f(x)= - 2/3(x+3)^2-6. The standard form of the equation is f(x)=2/3x^2+4x.

B.) The vertex form of the equation is f(x)=2/3(x+3)^2-6. The standard form of the equation is f(x)+-2/3x^2-4x.

C.) The vertex form of the equation is f(x)= - 2/3(x-3)^2-6. The standard form of the equation is f(x)=2/3x^2-4x.

D.) The vertex of the equation is f(x)=2/3(x+3)^2-6. The standard form of the equation is f(x)=2/3x^2+4x.

how 2 explain this?


Thanksss.