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Can you help answer these 2 math problems.

Can you help answer these 2 math problems.-example-1
User Faph
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Answer:

Question 7

Domain: (-2,3]

Range: (1,4]

Question 8

a) No

b) x=4 and x=-2

c) (0,3)

d) (3,0) and (-1,0)

Explanation:

Question 7

Domain refers to the span of x-values that a function encompasses. We can see from the graph that the lowest x value is -2 and the highest one is 3, and we also note the open circle at x=-2 (meaning x=-2 itself isn't included, which we note using curved brackets) and the closed circle at x=3 (meaning x=3 itself is included, which we note using square brackets). This leaves us with a domain of (-2,3]

Range meanwhile refers to the span of y-values that a function encompasses. We can see from the graph that the lowest y value is 1 and the highest one is 4, and we also note the open circle at y=1 (meaning y=1 itself isn't included, which we note using curved brackets) and the closed circle at y=4 (meaning y=4 itself is included, which we note using square brackets). This leaves us with a range of (1,4]

Question 8

a)

If the point (1,-3) is on the graph, then subbing 1 in for x and -3 in for f(x) should equate:

f(x)=x²-2x-3

-3=1²-2(1)-3?

-3=1-2-3?

-3=-4?

Since this doesn't equate, we can say the point is not on the line.

b)

The condition is that (x,5) is on the graph, so let's sub 5 in for f(x):

f(x)=x²-2x-3

5=x²-2x-3

x²-2x-8=0

(x-4)(x+2)=0

x=4 and x=-2

c)

The y-intercept is the point(s) at which the graph crosses the y-axis, ie where x=0, so to find it we sub 0 in for x:

f(x)=x²-2x-3

f(0)=0²-2(0)-3=-3

Therefore the y-intercept is (0,3)

d)

The c-intercept is the point(s) at which the graph crosses the x-axis, ie where y=0, so to find it we sub 0 in for f(x):

f(x)=x²-2x-3

0=x²-2x-3

(x-3)(x+1)=0

x=3 and x=-1

Therefore the x-intercepts are (3,0) and (-1,0)

User Walnutmon
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