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2 years ago

Graph the equation f(x)= 4(x)+5

Mathematics

2 answers:

faltersainse [42]2 years ago

5 0

Answer:

Graph is a straight line.

Step-by-step explanation:

We have given a function.

f(x) = 4x+5

We have to graph the function.

Sicce, f(x) = mx+c is equation of line where m is slope and c is y-intercept.

comparing given equation with equation of line, we have

m = 4 and c = 5.

putting f(x) = in given equation,we have

0 = 4x+5

x = -5/4 hence x-intercept = -5/4 = 1.25

Dram graph and point out the y-intercept and x-intercept.

Graph is attached.Graph is a straight line.

hoa [83]2 years ago

4 0

Answer:

Step-by-step explanation:

The 5 is the y intercept. So that point will be (0,5). The 4 in front of the x is the slope. SO from (0,5) you would up up 4 and over 1. This will bring you to (1,9). You can do this in both directions.

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3 years ago

The focus of a parabola is (−5,−1) and the directrix is y=−3. What is an equation of the parabola? (x+5)2=8(y+5)

Anna35 [415]

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(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
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2 years ago

Write an equation of the line that passes through each points (5,6) (3.4)

ioda

Answer:

$y = x + 1$ .

Step-by-step explanation:

Start by finding the slope of this line.

If a slanting line goes through $(x_{0},\, y_{0})$ and $(x_{1},\, y_{1})$ , where $x_{0} \ne x_{1}$ , the slope of this line would be:

$\begin{aligned}m &= \frac{y_{1} - y_{0}}{x_{1} - x_{0}}\end{aligned}$ .

The line in this question goes through $(5,\, 6)$ and $(3,\, 4)$ . Hence, the slope of this line would be:

$\begin{aligned}m &= \frac{y_{1} - y_{0}}{x_{1} - x_{0}} \\ &= \frac{6 - 4}{5 - 3} \\ &= 1\end{aligned}$ .

If a slanting line with a slope of $m$ and goes through the point $(x_{0},\, y_{0})$ , the equation of this line in the point-slope form would be:

$y - y_{0} = m,\, (x - x_{0})$ .

For the line in this question, the slope is $m = 1$ . Take $(5,\, 6)$ as the chosen point on this line. The point-slope form equation of this line would be:

$y - 6 = (x - 5)$ .

Rewrite to obtain the equation of this line in the slope-intercept form:

$y = x + 1$ .

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2 years ago

If u(x)=-2x²+3 and v(x)=1/x, what is the range of (u ° v)(x)?

Ludmilka [50]

$u(x)=-2x^2+3\\\\v(x)=\dfrac{1}{x}\\\\(u\ \circ\ v)(x)=-2\left(\dfrac{1}{x}\right)^2+3=-2\left(\dfrac{1}{x^2}\right)+3=-\dfrac{1}{x^2}+3\\\\\text{The range of}\ y=\dfrac{1}{x^2}\ \text{is all positive real numbers.}\\\\\text{The range of}\ y=-\dfrac{1}{x^2}\ \text{is all negative real numbers.}\\\\\text{The range of }\ (u\ \circ\ v)(x)=-\dfrac{1}{x^2}+3\ is\ (-\infty,\ 3)$

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3 years ago

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