Exercises — Intercepts — x-intercept, y-intercept

The second coordinate ($y$) is $0$. A point with $y=0$ lies on the x-axis (no up-down distance). So $(5,0)$ is an x-intercept. Picture: it sits on the flat horizontal road, $5$ steps to the right of the origin. See

The y-intercept is on the y-axis, where every point has $x=0$. So you set $==x=0==$. (The term $4x$ then vanishes.) Memory hook from the parent note: "Touch a road ⇒ the OTHER one is zero."

x-intercept — set $y=0$ (the $-4y$ term disappears): $3x=24⟹x=8.\text{Point }(8,0).$ y-intercept — set $x=0$ (the $3x$ term disappears): $-4y=24⟹y=-6.\text{Point }(0,-6).$ Notice the negative: solving $-4y=24$ means dividing by a negative, so $y$ comes out below the origin. That is correct — the line dips under the x-axis on the left. See

This is slope-intercept form $y=mx+c$ with $m=2$, $c=-9$. y-intercept: in $y=mx+c$ the constant $c$ is the y-intercept, because setting $x=0$ gives $y=c$. So $y$-intercept $=-9$, point $(0,-9)$. x-intercept: set $y=0$: $0=2x-9⟹2x=9⟹x=\frac{9}{2}=\mathrm{4.5.}\text{Point }(4.5,0).$

x-intercepts — set $y=0$; the crossings of the x-axis are the roots of the quadratic: $x^2-5x+4=0⟹(x-1)(x-4)=0⟹x=1\text{ or }x=4.$ Points $(1,0)$ and $(4,0)$. y-intercept — set $x=0$: $y=0-0+4=4.\text{Point }(0,4).$ The constant term $4$ is always the y-intercept of a polynomial, because every $x$-term vanishes at $x=0$.

We are handed the intercepts directly, so intercept form is the fast route: $\frac{x}{a}+\frac{y}{b}=1\text{with }a=6,b=-3.$ $\frac{x}{6}+\frac{y}{-3}=1.$ Multiply through by $6$ (to clear denominators): $x-2y=6.$ Check: $y=0\Rightarrow x=6$ ✓; $x=0\Rightarrow -2y=6\Rightarrow y=-3$ ✓.

The x-intercept comes from setting $y=0$: $2x+k\cdot 0=10⟹2x=10⟹x=5.$ This is true for every value of $k$, because the $ky$ term is multiplied by $0$ and disappears. So the x-intercept being $5$ tells us nothing about $k$ — $k$ cannot be determined. Any $k\ne 0$ works. Lesson: the x-intercept only "sees" the $x$-coefficient and the constant, never the $y$-coefficient.

Set $y=0$: $x^2+2x+5=0$. Use the quadratic-root discriminant $b^2-4ac$: $\Delta =2^2-4(1)(5)=4-20=-16<0.$ A negative discriminant means no real roots, so the parabola has no x-intercepts — it never touches the x-axis. Geometrically the whole curve floats above the x-axis (its lowest point is at $y=5-1=4>0$). Its y-intercept still exists: at $x=0$, $y=5$.

Intercepts first.

• $P$: set $y=0\Rightarrow x=8\Rightarrow P=(8,0)$.
• $Q$: set $x=0\Rightarrow 2y=8\Rightarrow y=4\Rightarrow Q=(0,4)$.

These two intercepts plus the origin form a right triangle, with the two legs lying along the axes (they meet at the right angle at $O$). See

• Base (along x-axis) $=8$ (the distance from $O$ to $P$, a coordinate distance).
• Height (along y-axis) $=4$. $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 8\times 4=16.$

Passing through $(0,6)$ means the y-intercept is $6$, so slope-intercept form gives: $y=-\frac{3}{2}x+6.$ x-intercept — set $y=0$: $0=-\frac{3}{2}x+6⟹\frac{3}{2}x=6⟹x=4.$ So $a=4$, $b=6$. Intercept form: $\frac{x}{4}+\frac{y}{6}=1.$ Check: multiply by $12$: $3x+2y=12$; rearrange $\Rightarrow y=-\frac{3}{2}x+6$ ✓.

x-intercept means set $y=0$ in each line separately.

• Line 1: $2x+0=6\Rightarrow x=3\Rightarrow (3,0)$.
• Line 2: $x-0=3\Rightarrow x=3\Rightarrow (3,0)$.

Both x-intercepts are $(3,0)$ — the same point. So the two lines actually cross the x-axis at the same place, which means $(3,0)$ is their intersection point too. (You can confirm: $(3,0)$ satisfies both equations.)

Find the intercepts:

• x-intercept: $0=3x\Rightarrow x=0\Rightarrow (0,0)$.
• y-intercept: $y=3(0)=0\Rightarrow (0,0)$.

Both intercepts are $0$ — the line passes through the origin. Intercept form needs $a\ne 0$ and $b\ne 0$ (you divide by them), so plugging $a=0,b=0$ would mean dividing by zero, which is undefined. Therefore $y=3x$ cannot be written in intercept form. Its single "crossing" of both axes is the origin, and you just describe it as $y=3x$.

(a) $y=4$. This is a flat horizontal line, always $4$ above the x-axis.

• y-intercept: set $x=0\Rightarrow y=4\Rightarrow (0,4)$. ✓ Exists.
• x-intercept: set $y=0$. But the line insists $y=4$ everywhere, so $y=0$ is never true. It runs parallel to the x-axis and never touches it — no x-intercept.

(b) $x=-2$. A vertical line, always $2$ to the left.

• x-intercept: set $y=0$; the line already allows any $y$, so at $y=0$, $x=-2\Rightarrow (-2,0)$. ✓ Exists.
• y-intercept: set $x=0$. But the line insists $x=-2$, never $0$. Parallel to the y-axis ⇒ no y-intercept.

Equal intercepts $a=b$ put the line in intercept form: $\frac{x}{a}+\frac{y}{a}=1⟹x+y=a.$ Now use the point $(3,5)$ — it must satisfy the equation: $3+5=a⟹a=8.$ So the equation is $x+y=8$. Check intercepts: $y=0\Rightarrow x=8$; $x=0\Rightarrow y=8$ — equal ✓. And $(3,5)$: $3+5=8$ ✓.

Rewrite: $\frac{x}{a}+\frac{y}{4}=1$. As $a\to \infty$, the term $\frac{x}{a}\to 0$ for any fixed $x$ (dividing by a huge number). What remains is: $\frac{y}{4}=1⟹y=4.$ So the line flattens into the horizontal line $y=4$. Its x-intercept $a$ has run off to infinity — meaning the line no longer crosses the x-axis at any finite point. This matches Exercise 5.2(a): a horizontal line has no (finite) x-intercept. The "$a\to \infty$" is the precise limiting picture of losing the x-intercept.

Set the RIGHT variable to zero (name after the axis) ::: x-intercept → $y=0$; y-intercept → $x=0$. Handle a negative intercept correctly ::: divide including the sign; the sign tells you which side of the origin. A quadratic with $\Delta <0$ has how many x-intercepts? ::: Zero — the curve never meets the x-axis. Area of the axis triangle from intercepts $a,b$ ::: $\frac{1}{2}|a||b|$ (absolute values). Why can't $y=3x$ use intercept form? ::: Both intercepts are $0$ ⇒ dividing by zero. Intercepts of a horizontal line $y=k$ ($k\ne 0$) ::: y-intercept $(0,k)$; NO x-intercept.