Slope for square root equation

[andreo]

Slope for square root equation
i have this equation :
y=ax^1/2 + bx; and have one of (x,y) point is (4,16)
I want my equation above have same slope as y=2x^1/2 + x,
What are a and b value in my first equation above?

 

[stapel]

Slope for square root equation
i have this equation :
y=ax^1/2 + bx; and have one of (x,y) point is (4,16)
I want my equation above have same slope as y=2x^1/2 + x,
What are a and b value in my first equation above?

When one is working with "slopes" of non-straight lines, one is usually working with calculus. Are you? If you are actually working in pre-calculus algebra (the category to which you have posted), what method(s) or algorithm(s) have you been given for this? How far have you gotten in working this?

Please be complete. Thank you!

 

[andreo]

square root equation

y=ax^1/2 + bx; this equation have point at (x,y) = (16,8)
i want my equation above have slope same as this equation y=2x^1/2 + x;
could you determine what is a and b value in my first equation above?

 

[andreo]

When one is working with "slopes" of non-straight lines, one is usually working with calculus. Are you? If you are actually working in pre-calculus algebra (the category to which you have posted), what method(s) or algorithm(s) have you been given for this? How far have you gotten in working this?

Please be complete. Thank you!

Thank you for reply my question,

Actually i'm a telecommunication engineer and work with some equation like insertion loss cable that have formula y=ax^1/2 + bx,with a and b is constant, y is insertion loss cable and x is frequency .
Specification cable have insertion loss graphic and equation y=2x^1/2 + x; in that equationwhen x = 16, y=24 (16,24),
I do measurement on cable and i have result in x=16 the y=8 (16,8) so I would like to know the value a and b in y=ax^1/2 + bx with x=16 and y=8?

 

[stapel]

...i...work with...equation...that have formula y=ax^1/2 + bx, with a and b is constant....
I would like to know the value a and b in y=ax^1/2 + bx with x=16 and y=8?

You have one equation in two unknowns; it is not possible to solve this for the values of those two unknowns.

 

[JeffM]

Thank you for reply my question,

Actually i'm a telecommunication engineer and work with some equation like insertion loss cable that have formula y=ax^1/2 + bx,with a and b is constant, y is insertion loss cable and x is frequency .
Specification cable have insertion loss graphic and equation y=2x^1/2 + x; in that equationwhen x = 16, y=24 (16,24),
I do measurement on cable and i have result in x=16 the y=8 (16,8) so I would like to know the value a and b in y=ax^1/2 + bx with x=16 and y=8?

As staple explained, with one data point, it is impossible to determine two values. You need to test at an ADDITIONAL frequency.

First test: x = p and y = q. Second test: x = r and y = s.

\(\displaystyle q = a\sqrt{p} + bp \implies b = \dfrac{q - a\sqrt{p}}{p}.\)

\(\displaystyle s = a\sqrt{r} + br \implies s = a\sqrt{r} + \dfrac{qr - ar\sqrt{p}}{p} \implies\)

\(\displaystyle ps = ap\sqrt{r} + qr - ar\sqrt{p} \implies ps - qr = a\left(p\sqrt{r} - r\sqrt{p}\right) \implies\)

\(\displaystyle a = \dfrac{ps - qr}{p\sqrt{r} - r\sqrt{p}}\).

Compute a from test results, then b.