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Chapter 2: Problem 57

Find the \(x\) - and \(y\) -intercepts. $$ x^{2}+y=9 $$

Short Answer

Expert verified
x-intercepts: (3, 0) and (-3, 0); y-intercept: (0, 9).

Step by step solution

01

Find the x-intercepts

To find the x-intercepts, set y to 0 in the equation and solve for x. \[ x^2 + 0 = 9 \] This simplifies to: \[ x^2 = 9 \] Taking the square root of both sides: \[ x = \pm 3 \] So, the x-intercepts are at \((3, 0)\) and \((-3, 0)\).
02

Find the y-intercept

To find the y-intercept, set x to 0 in the equation and solve for y. \[ 0^2 + y = 9 \] This simplifies to: \[ y = 9 \] So, the y-intercept is at \((0, 9)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

x-intercepts
To understand x-intercepts, you need to know that they are points where a graph crosses the x-axis. At these points, the y-value is always zero. For the given equation, \(x^2 + y = 9\), to find the x-intercepts, you set y to 0 and solve for x. Here is how:

\ \( x^2 + 0 = 9 \) simplifies to \( x^2 = 9 \).

Taking the square root of both sides, we get \( x = \pm 3 \).

Thus, the x-intercepts are at \((3, 0)\) and \((-3, 0)\). These points mean that when the value of y is zero, x can be either 3 or -3. Now you know how to find x-intercepts!
y-intercepts
Y-intercepts are the points where a graph crosses the y-axis. At these points, the x-value is always zero. For the given equation, \(x^2 + y = 9\), to find the y-intercept, you set x to 0 and solve for y. Here is how:

\ \( 0^2 + y = 9 \) simplifies to \( y = 9 \).

Thus, the y-intercept is at \((0, 9)\). This means that when the value of x is zero, y is 9. Finding y-intercepts is simple once you understand the method. Always set x to 0 and solve for y.
solving quadratic equations
Solving quadratic equations is a key skill in algebra. A quadratic equation is usually in the form \( ax^2 + bx + c = 0 \). The given equation \( x^2 + y = 9 \) is a simple quadratic in terms of x.

Here are common methods to solve quadratics:
  • Factoring: Find two numbers that multiply to give \( ac \) and add to give \( b \).
  • Quadratic Formula: Use the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Completing the Square: Make the quadratic into a perfect square trinomial, then solve.


In the given equation, setting y to 0 changes it to \( x^2 = 9 \). Solving, you find \( x = \pm 3 \). This straightforward case shows the essence of solving quadratics: isolate the term with the square and simplify. Master these methods to tackle any quadratic equation with ease.

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