At what ##x## value is the tangent equally inclined to the given curve?

[chwala]

Homework Statement

See attached.

Relevant Equations

differentiation.

I had to look this up; will need to read on it.

from my research,

https://byjus.com/question-answer/t...om-the-points-1-2-and-3-4-is-ax-by-c-0-where/

...
I have noted that at equally inclined; the slope value is $1$.

$\dfrac{dy}{dx} = 2x^2+x=1$

$2x^2+x-1=0$

$x=-1$ or $x=0.5$

the steps are clear; but i need to understand the concept...i guess more reading on my part. ...just sharing in the event one has insight to offer. Cheers.

 

[Mark44]

Homework Statement: See attached.
Relevant Equations: differentiation.

I had to look this up; will need to read on it.

View attachment 342013

from my research,

https://byjus.com/question-answer/t...om-the-points-1-2-and-3-4-is-ax-by-c-0-where/

...
I have noted that at equally inclined; the slope value is $1$.

$\dfrac{dy}{dx} = 2x^2+x=1$

$2x^2+x-1=0$

$x=-1$ or $x=0.5$

the steps are clear; but i need to understand the concept...i guess more reading on my part. ...just sharing in the event one has insight to offer. Cheers.

It's pretty straightforward. If you graph $y = \frac 2 3 x^3 + \frac 1 2 x^2$, there are points in the first and third quadrants at which the slope is 1, namely, at (-1, -1/6) and (1/2, 5/24).

 

[chwala]

Noted, in general can we also have equally inclined when slope $=-1$?

 

[pasmith]

Noted, in general can we also have equally inclined when slope $=-1$?

I would say yes.

 

[Orodruin]

Just for reference, here is the plot of the function and the line $y = x$.

Pretty suggestive that at x = -1 and 0.5 are the slope is indeed the same slope as $y = x$.

 

[Orodruin]

Noted, in general can we also have equally inclined when slope $=-1$?

In general, yes. For this particular case, it never happens as
$$
2x^2 + x + 1 = 0
$$
has roots $-1/4 \pm \sqrt{1/16 - 1/2} = -1/4 \pm i \sqrt{7}/4$, which both have non-zero imaginary part.