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Mathematics

1 answer:

givi [52]3 years ago

5 0

Answer:

a. y = 0

b. x = 0

c. y = x

d. y = -x

Step-by-step explanation:

a. Since it is a horizontal line, its gradient does not change so its gradient shall remain as 0. Thus, y = 0x which is basically y = 0.

b. Since it is a vertical line, it is impossible to define its gradient or y-intercept but its x-intercept would always be at 0 so x = 0.

c. Since it is a line that is gradually getting higher across the x-axis, its gradient would be 1. It's y-intercept is 0 so the equation of this line would be y = 1x + 0, which is basically y = x. (We don't write 1x, but rather just x, putting zeros behind an equation is redundant)

d. Since it is a line that is gradually getting lower across the x-axis, its gradient would be -1. It's y-intercept is also 0 so the equation of this line would be y = -1x + 0, which is basically y = -x ((We don't write -1x, but rather just -x, putting zeros behind an equation is redundant)

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prisoha [69]

Answer:

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Step-by-step explanation:

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Answer:

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3 years ago

The graph of which function will have a maximum and a y-intercept of 4? f(x) = 4x2 + 6x – 1 f(x) = –4x2 + 8x + 5 f(x) = –x2 + 2x

V125BC [204]

Answer:

Option C (f(x) = $-x^2 + 2x + 4$ )

Step-by-step explanation:

In this question, the first step is to write the general form of the quadratic equation, which is f(x) = $ax^2 + bx + c$ , where a, b, and c are the arbitrary constants. There are certain characteristics of the values of a, b, and c which determine the nature of the function. If a is a positive coefficient (i.e. if a>0), then the quadratic function is a minimizing function. On the other hand, a is negative (i.e. if a<0), then the quadratic function is a maximizing function. Since the latter condition is required, therefore, the first option (f(x) = $4x^2 + 6x - 1$ ) and the last option (f(x) = $x^2 + 4x - 4$ ) are incorrect. The features of the values of b are irrelevant in this question, so that will not be discussed here. The value of c is actually the y-intercept of the quadratic equation. Since the y-intercept is 4, the correct choice for this question will be Option C (f(x) = $-x^2 + 2x + 4$ ). In short, Option C fulfills both the criteria of the function which has a maximum and a y-intercept of 4!!!