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erastovalidia [21]
3 years ago
13

Which function is graphed? A f(x)=x−1 B f(x)=−1 C f(x)=1 D f(x)=x+1

Mathematics
2 answers:
Anit [1.1K]3 years ago
5 0

The answer is B.

f(x) = x-1 will give you a graph of linear function also known as "line"

f(x) = -1 will give you the exact graph like the picture. It's the graph of constant function.

f(x) = 1 will give you a graph of constant function like f(x) = -1. However the graph passes (0,-1) , not (0,1)

f(x) = x+1 will give you a graph of linear function.

Linear Function doesn't give you either vertical or horizontal line except for constant one.

JulsSmile [24]3 years ago
3 0

Answer:

It is b

Step-by-step explanation:

I graphed all of them

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Answer:

g = 118°, h = 62°, k = 140°, m = 40°

Step-by-step explanation:

g = 118°

h = 180 - 118 = 62°

k = 140°

m = 180 - 140 = 40°

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Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
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Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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