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?

Jaime Coleman
2022-04-06
Answered

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?

You can still ask an expert for help

Madalynn Acosta

Answered 2022-04-07
Author has **11** answers

The y-intercept of a function $f(x)$ is the point where the function intersects the y-axis (if in fact it does intersect the y-axis) and it is found by evaluating $f(0)$, i.e., finding the value of $f(x)$ when $x=0$.

For example,

- The line $f(x)=y=3x+2$ intersects the x-axis at when $x=0:\phantom{\rule{thickmathspace}{0ex}}$:when $\phantom{\rule{thickmathspace}{0ex}}y=3\cdot 0+2=2$. This is the function's y-intercept.

- The parabola $f(x)=y=2{x}^{2}+8$ intersects the y-axis when $x=0:\phantom{\rule{thickmathspace}{0ex}}$:when $\phantom{\rule{thickmathspace}{0ex}}y=2(0{)}^{2}+8=8$. This is the function's y-intercept.

For example,

- The line $f(x)=y=3x+2$ intersects the x-axis at when $x=0:\phantom{\rule{thickmathspace}{0ex}}$:when $\phantom{\rule{thickmathspace}{0ex}}y=3\cdot 0+2=2$. This is the function's y-intercept.

- The parabola $f(x)=y=2{x}^{2}+8$ intersects the y-axis when $x=0:\phantom{\rule{thickmathspace}{0ex}}$:when $\phantom{\rule{thickmathspace}{0ex}}y=2(0{)}^{2}+8=8$. This is the function's y-intercept.

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 2022-05-07

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.

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.

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 2022-05-09

The gradient is $m=-4/3$, and the point given is $(3/2,3)$

Let $c$ be the $y$-intercept

$y=mx+c$

$y=(-4/3)x+c$

If I substitute $x$ and $y$ from the point given, $3=(-4/3)(3/2)+c$ and look for $c$, I get $c=5$

But according to my answer sheet, $c=33/8$ by using $y-{y}_{1}=m(x-{x}_{1})$

Why are the two methods giving different answers?

Let $c$ be the $y$-intercept

$y=mx+c$

$y=(-4/3)x+c$

If I substitute $x$ and $y$ from the point given, $3=(-4/3)(3/2)+c$ and look for $c$, I get $c=5$

But according to my answer sheet, $c=33/8$ by using $y-{y}_{1}=m(x-{x}_{1})$

Why are the two methods giving different answers?

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 2022-05-09

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?