What is the formal or general term for the y-intercept? Is there a general term for the value at which a function intercepts the vertical axis, in the Cartesian plane?

Ariel Wilkinson
2022-10-05
Answered

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Corbin Hanson

Answered 2022-10-06
Author has **10** answers

You do not need to associate the symbol, "$y$", as a function of $x$. Instead, you could just write $f(x)=\cdots $ or $f(x,y,z,\cdots )=\cdots $, if the function contains multiple variables. This means your vertical axis become the values for $f(x)$ and not "$y$" anymore. Hence, you describe the intercept at the vertical axis, mathematically, where the $x=0$ or similar if your function has more variables.

Alternatively, if you are really keen on describing the intercept based relative cartesian plane (no variables involved), it may be best to address it as the "vertical intercept".

Alternatively, if you are really keen on describing the intercept based relative cartesian plane (no variables involved), it may be best to address it as the "vertical intercept".

asked 2022-06-14

Consider the surface given by the function $f(x,y)=\frac{4}{\sqrt{{x}^{2}+{y}^{2}-9}}$. What is the range of $f$? Determine any $y$-intercept(s) of $f$.

For range, I realize that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number, so long as ${x}^{2}+{y}^{2}-9$ is positive or equal to zero. However, when looking at $f$, we know that the denominator cannot equal zero, which means that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number greater than $0$. Would this mean that the range of $f$ is $(0,\mathrm{\infty})$?

For the $y$-intercept, I know that both $x$ and $z$ need to equal zero. Would this mean that there are no $y$-intercepts because $0$ is not in the range of $f$?

For range, I realize that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number, so long as ${x}^{2}+{y}^{2}-9$ is positive or equal to zero. However, when looking at $f$, we know that the denominator cannot equal zero, which means that $\sqrt{{x}^{2}+{y}^{2}-9}$ can be any number greater than $0$. Would this mean that the range of $f$ is $(0,\mathrm{\infty})$?

For the $y$-intercept, I know that both $x$ and $z$ need to equal zero. Would this mean that there are no $y$-intercepts because $0$ is not in the range of $f$?

asked 2022-06-23

Need help with the steps of finding the slope and $y$-intercept of this equation: $f(x)=3x-\frac{1}{5}$

asked 2022-05-09

An equation to the tangent line of $h(x)=\mathrm{tan}(x)+\mathrm{cos}(x)$ is given by $y=mx+b$ where $m$ is the slope and $b$ is the $y$-intercept. If $x=\frac{\pi}{4}$ find $m$.

asked 2022-05-31

y=0.5x-2

asked 2022-05-09

Find the $y$-intercept of the curve that passes through the point $(2,1)$ with the slope at $(x,y)$ of $\frac{-9}{{y}^{2}}$

$\frac{dy}{dx}=\frac{-9}{{y}^{2}}$

$\int {y}^{2}dy=\int -9dx$

$\frac{{y}^{3}}{3}=-9x+{C}_{1}$

${y}^{3}=-27x+C$ $(C=3{C}_{1})$

$y=(-27x+C{)}^{1/3}$

$1=(-27(2)+C{)}^{1/3}$

$C=54$

$0=(-27x+54{)}^{1/3}$

$y-intercept=(-2,0)$

Where is mistake?

$\frac{dy}{dx}=\frac{-9}{{y}^{2}}$

$\int {y}^{2}dy=\int -9dx$

$\frac{{y}^{3}}{3}=-9x+{C}_{1}$

${y}^{3}=-27x+C$ $(C=3{C}_{1})$

$y=(-27x+C{)}^{1/3}$

$1=(-27(2)+C{)}^{1/3}$

$C=54$

$0=(-27x+54{)}^{1/3}$

$y-intercept=(-2,0)$

Where is mistake?

asked 2022-06-26

What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11?

asked 2022-07-05

Why is the y intercept for this equation positive?

Rational function

$\frac{(x+1)(3x+2)}{{(x+1)}^{2}}$

I have found that the two asymptotes are: $y=3$ and $x=-1$ and there are no turning points for this function.

To find the $x$ intercept:

$0=\frac{(x+1)(3x+2)}{{(x+1)}^{2}}$

Hence

$3{x}^{2}-x-2=0$

${x}_{1}=1\text{}\text{}\text{}{x}_{2}=-\frac{2}{3}$

Simularly to find the $y$ intercept I did the following:

$\frac{(0+1)(3(0)+2)}{{(0+1)}^{2}}$

$y=-2$

However plotting this into a graphing software shows that the $y$ intercept is $y=2$ and $x$ intercept is only $x=-\frac{2}{3}$ and not $x=1$.

Why are the intercepts incorrect?

Rational function

$\frac{(x+1)(3x+2)}{{(x+1)}^{2}}$

I have found that the two asymptotes are: $y=3$ and $x=-1$ and there are no turning points for this function.

To find the $x$ intercept:

$0=\frac{(x+1)(3x+2)}{{(x+1)}^{2}}$

Hence

$3{x}^{2}-x-2=0$

${x}_{1}=1\text{}\text{}\text{}{x}_{2}=-\frac{2}{3}$

Simularly to find the $y$ intercept I did the following:

$\frac{(0+1)(3(0)+2)}{{(0+1)}^{2}}$

$y=-2$

However plotting this into a graphing software shows that the $y$ intercept is $y=2$ and $x$ intercept is only $x=-\frac{2}{3}$ and not $x=1$.

Why are the intercepts incorrect?