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3 years ago

How do I graph x-y=6

1 answer:

6 0

You change the equation into a y=mx+b form so it would be y=x-6 and so when you graph it, go down six on the y-axis and make a straight line that goes diagonally through the boxes. Not sure if you were asking how to change the equation to graphing form or how to actually graph the line but hopefully this helps. :)

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When the smaller of two consecutive integers is added to two times the larger, the result is 35. Find the integers.

Aleonysh [2.5K]

x+2(x+1)=35

x+2x+2=35

3x=35-2=33

x=33/3=11

smaller number=11

larger number=11+1=12

4 0

3 years ago

Im stuck .... What is the solution for the equation 2(3x−7)=−4(x+5)+3? plss show work

Ipatiy [6.2K]

Answer:

$2(3x - 7) = - 4(x + 5) + 3 \\ 6x - 14 = - 4x - 20 + 3 \\ 6x + 4x = - 20 + 3 + 14 \\ 10x = - 3 \\ \\ x = \frac{ - 3}{10} \\ \\ x = - 0.3$

I hope I helped you^_^

5 0

3 years ago

What would y=x^2 +x+ 2 be in vertex form

balu736 [363]

Answer:

$y = (x + \frac{1}{2} )^{2} + \frac{7}{4}$

Step-by-step explanation:

$y = {x}^{2} + x + 2$

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

$y = a(x - h)^{2} + k$

The following equation above is the vertex form of Quadratic Function.

Vertex — Formula

$h = - \frac{b}{2a} \\ k = \frac{4ac - {b}^{2} }{4a}$

We substitute the value of these terms from the standard form.

$y = a {x}^{2} + bx + c$

$h = - \frac{1}{2(1)} \\ h = - \frac{ 1}{2}$

Our h is - 1/2

$k = \frac{4(1)(2) - ( {1})^{2} }{4(1)} \\ k = \frac{8 - 1}{4} \\ k = \frac{7}{4}$

Our k is 7/4.

Vertex — Calculus

We can use differential or derivative to find the vertex as well.

$f(x) = a {x}^{n}$

Therefore our derivative of f(x) —

$f'(x) = n \times a {x}^{n - 1}$

From the standard form of the given equation.

$y = {x}^{2} + x + 2$

Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'

$f'(x) = (2 \times 1 {x}^{2 - 1} ) + (1 \times {x}^{1 - 1} ) + 0$

Any constants that are differentiated will automatically become 0.

$f'(x) = 2 {x}+ 1$

Then we substitute f'(x) = 0

$0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - \frac{1}{2}$

Because x = h. Therefore, h = - 1/2

Then substitute x = -1/2 in the function (not differentiated function)

$y = {x}^{2} + x + 2$

$y = ( - \frac{1}{2} )^{2} + ( - \frac{1}{2} ) + 2 \\ y = \frac{1}{4} - \frac{1}{2} + 2 \\ y = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} \\ y = \frac{7}{4}$