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Chapter 1: Problem 32

Find the \(x\) - and \(y\) -intercepts of the graph of the equation. $$y^{2}=x+1$$

Short Answer

Expert verified
The x-intercept is \(-1, 0\) and y-intercepts are \(0, 1\) and \(0, -1\).

Step by step solution

01

Find the x-intercepts

To find the x-intercepts, set \(y = 0\). Substitute 0 for y in the equation to get \(0^{2} = x+1\). Simplify to get \(x +1=0\), or \(x=-1\). So the x-intercept of the graph of the equation are \(-1, 0\).
02

Find the y-intercepts

To find the y-intercepts, set \(x = 0\). Substitute 0 for x in the equation to get \(y^{2} = 0+1\). Solve for y to get \(y = \sqrt{1}\)or \(y=-\sqrt{1}\), which simplifies to \(y = 1\) or \(y = -1\). The y-intercepts of the graph of the equation are \(0, 1\) and \(0, -1\).

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Key Concepts

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

x-intercepts
When dealing with graphs, the x-intercepts are crucial points where the graph crosses the x-axis. These are the points where the output of your equation or function is zero. To find the x-intercepts from an equation, you simply set the y-value to zero and solve for x.
For instance, given the equation \(y^2 = x + 1\), we substitute \(y = 0\) into this equation:
  • Initial equation: \(y^2 = x + 1\).
  • Substitute \(y = 0\): \(0^2 = x + 1\).
  • Simplify to find \(x\): \(x + 1 = 0\).
  • Solve for \(x\): \(x = -1\).
Thus, the x-intercept is at the point \((-1, 0)\). It's essential to understand that this means at this point, the graph of the equation touches the x-axis, with a corresponding y-value of zero.
y-intercepts
The y-intercepts represent the points where the graph crosses the y-axis. These are the locations where x is zero, showing the direct influence of other terms in the equation. To find the y-intercepts, you set \(x = 0\) and solve for y.
Let鈥檚 consider the same equation: \(y^2 = x + 1\). Set \(x = 0\) in the equation to find:
  • Initial equation: \(y^2 = x + 1\).
  • Substitute \(x = 0\): \(y^2 = 0 + 1\).
  • This simplifies to \(y^2 = 1\).
  • Solve for \(y\): \(y = \sqrt{1}\) or \(y = -\sqrt{1}\).
Therefore, the y-intercepts are at points \((0, 1)\) and \((0, -1)\). Here, the graph intersects the y-axis at two distinct points due to the nature of this quadratic relationship in the original equation.
quadratic equations
Quadratic equations typically appear in the form \(ax^2 + bx + c = 0\), involving the variable squared. However, equations like \(y^2 = x + 1\) can also relate to quadratic characteristics when one side is considered a function of another. Quadratic equations can graphically form parabolas, which have properties that include vertex points, as well as x and y-intercepts.
When identifying the intercepts or solving quadratic equations:
  • x-intercepts are found by setting the equation equal to zero and solving for x.
  • y-intercepts are discovered by setting x to zero and resolving the equation.
  • They often yield two solutions because of the squared term, leading to the two intercepts as seen in the given example for y-intercepts: \(0, 1\) and \(0, -1\).
Understanding these concepts is key to mastering how changes in quadratic equations impact their graphs. Whether you're reading an equation or graphing it, these skills will allow you to deeply understand and predict the graph鈥檚 behavior.

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