Need to find the $y$-intercept of the tangent to $y=\frac{18}{{x}^{2}+2}$ at point $(1,6)$.

I keep getting $0$, but I don't think that's right.

I keep getting $0$, but I don't think that's right.

dresu9dnjn
2022-05-07
Answered

Need to find the $y$-intercept of the tangent to $y=\frac{18}{{x}^{2}+2}$ at point $(1,6)$.

I keep getting $0$, but I don't think that's right.

I keep getting $0$, but I don't think that's right.

You can still ask an expert for help

Corinne Choi

Answered 2022-05-08
Author has **10** answers

Step 1: Compute the tangent line.

- The tangent line is a line passing through the point $(1,6)$ with the same slope as the curve that that point.

- In order to write down a line, you need a point on the line and the slope of the line. You already have a point, but you need to find the slope of the line.

- The slope of the line is the derivative at the point $(1,6)$, since the function is

$\frac{18}{{x}^{2}+2},$

its derivative is

$-\frac{36x}{({x}^{2}+2{)}^{2}}$

plugging in $x=1$ from the point $(1,6)$, we get that the slope is

$-\frac{36}{{3}^{2}}=-4.$

This is the slope of the line of interest.

- Therefore, using the point-slope form of a line, you get that the tangent line is $y-6=(-4)(x-1)$.

Step 2: Find the $y$-intercept.

- You have the point-slope form of a line, to get $y-6=-4(x-1)$. We can turn this into slope-intercept form to get $y=-4x+10$.

- Therefore, your $y$-intercept is $10$.

- The tangent line is a line passing through the point $(1,6)$ with the same slope as the curve that that point.

- In order to write down a line, you need a point on the line and the slope of the line. You already have a point, but you need to find the slope of the line.

- The slope of the line is the derivative at the point $(1,6)$, since the function is

$\frac{18}{{x}^{2}+2},$

its derivative is

$-\frac{36x}{({x}^{2}+2{)}^{2}}$

plugging in $x=1$ from the point $(1,6)$, we get that the slope is

$-\frac{36}{{3}^{2}}=-4.$

This is the slope of the line of interest.

- Therefore, using the point-slope form of a line, you get that the tangent line is $y-6=(-4)(x-1)$.

Step 2: Find the $y$-intercept.

- You have the point-slope form of a line, to get $y-6=-4(x-1)$. We can turn this into slope-intercept form to get $y=-4x+10$.

- Therefore, your $y$-intercept is $10$.

studovnaem4z6

Answered 2022-05-09
Author has **3** answers

First you find the slope of tangent line at the point of tangency. Slope of tangent line is derivative of the function evaluated at the point of tangency. For $y=18/({x}^{2}+2)$ we use quotient rule to get ${y}^{\prime}=-36x/({x}^{2}+2{)}^{2}$. $m={y}^{\prime}(1)=-4$ is the slope and $y=6-4(x-1)$ is the equation of the tangent line. To find the y intercept we let $x=0$ and get $y=10.$ Thus the point $(0,10)$ is the $y$ intercept of the tangent line.

asked 2022-05-15

This is equation of a curve $\sqrt{y}+\sqrt{x}=\sqrt{A}$

$A$ is a positive constant

$T$ is a tangent of the curve from any point on it

$B$ is the y-intercept of $T$

$C$ is the x-intercept of $T$

Prove that $B+C=A.$.

$A$ is a positive constant

$T$ is a tangent of the curve from any point on it

$B$ is the y-intercept of $T$

$C$ is the x-intercept of $T$

Prove that $B+C=A.$.

asked 2022-05-10

Given $f(x)=\frac{1}{4}(x+4{)}^{2}-2$

Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry

What I have so far:

Vertex:$(-4,-2)$

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

$x$-intercept: $2$

Axis of symmetry $x=-4$

Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry

What I have so far:

Vertex:$(-4,-2)$

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

$x$-intercept: $2$

Axis of symmetry $x=-4$

asked 2022-04-06

how to draw functions on a coordinate plane, and he mentioned something about the Y-Intercept being an important step in creating/solving a function. But, what exactly is a Y-Intercept?

asked 2022-04-06

Let $f(x)={x}^{2}+x-6$.

a. Write down the $y$-intercept of the graph of $f$.

how do we figure this out? I know $f(x)$ means $y$, so do we use the quadratic formula? ${x}^{2}+x-6$.

b. Solve $f(x)=0$.

We plug $0$ into the $x$’s in the original formula, right?

a. Write down the $y$-intercept of the graph of $f$.

how do we figure this out? I know $f(x)$ means $y$, so do we use the quadratic formula? ${x}^{2}+x-6$.

b. Solve $f(x)=0$.

We plug $0$ into the $x$’s in the original formula, right?

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 2022-05-09

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 2022-05-08

Two questions involving quadratics. Next to a diagram of a parabola with a ma point with the y intercept of $(0,p)$ and roots $(-1,0)$ and $(p,0)$ it says:3a) Show that the equation of the parabola is $y=p+(p-1)x-{x}^{2}$

3b) For what value of $p$ will the line $y=x+p$ be a tangent to this curve.

3b) For what value of $p$ will the line $y=x+p$ be a tangent to this curve.