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Chapter 11: Problem 27

Sketch the graph of the given equation. Find the intercepts; approximate to the nearest tenth where necessary. $$y=x^{2}+3 x+2$$

Short Answer

Expert verified
The intercepts are (0,2), (-1,0), and (-2,0). The vertex is (-1.5,-0.25).

Step by step solution

01

Identify the type of equation

The given equation is a quadratic equation of the form: \(y = ax^2 + bx + c \) For the given equation, a = 1, b = 3, and c = 2.
02

Find the y-intercept

To find the y-intercept, set \(x=0\) and solve for \(y\): \( y = 0^2 + 3(0) + 2 = 2 \) Therefore, the y-intercept is at (0, 2).
03

Find the x-intercepts

To find the x-intercepts, set \(y=0\) and solve for \(x\):\( 0 = x^2 + 3x + 2\) Factorize the quadratic equation: \((x + 1)(x + 2) = 0\) Solve for \(x\): Hence, \(x = -1\) and \(x = -2\). Therefore, the x-intercepts are at (-1,0) and (-2,0).
04

Determine the vertex

The vertex form of a quadratic equation is: \( y = a(x-h)^2 + k \) where \( (h, k) \) is the vertex. The vertex for \(y = x^2 + 3x + 2\) can be found using \( h = -\frac{b}{2a} \): \( h = -\frac{3}{2(1)} = -1.5 \) Now, substitute \(x = -1.5\) into the original equation to find \(y\): \( y = (-1.5)^2 + 3(-1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 \) So the vertex is at (-1.5, -0.25).
05

Sketch the graph

Now that the intercepts and vertex are known, plot the points: (0,2), (-1,0), (-2,0), and (-1.5, -0.25) on a graph. Draw the parabolic curve passing through these points. The parabola opens upwards since the coefficient of \(x^2\) is positive.

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Key Concepts

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

Y-Intercept
To understand the y-intercept, we need to consider where the graph crosses the y-axis. This happens when the value of x is zero.
For our equation, we substitute x=0 into the equation:
\( y = 0^2 + 3(0) + 2 = 2 \)
The y-intercept is at the point (0, 2). This means the graph of the equation will cross the y-axis at 2.
Always remember:
  • Set \( x = 0 \) to find the y-intercept.
X-Intercepts
X-intercepts occur where the graph crosses the x-axis, meaning the value of y is zero.
To find the x-intercepts, we set y=0 and solve for x:
\( 0 = x^2 + 3x + 2 \)
We factorize the equation:
\((x + 1)(x + 2) = 0 \)
This gives us two solutions:
\( x = -1 \) and \( x = -2 \)
The x-intercepts are (-1, 0) and (-2, 0).
Quick tips for finding x-intercepts:
  • Set \( y = 0 \) and solve for x.
  • Factorize if possible.
Vertex Form
The vertex form of a quadratic equation helps us easily find the vertex, the highest or lowest point of the parabola.
The standard vertex form is:
\( y = a(x-h)^2 + k \)
where \( (h,k) \) is the vertex.
For the given equation \( y = x^2 + 3x + 2 \), we use the formula:
\( h = -\frac{b}{2a} \)
Substituting a=1, b=3:
\( h = -\frac{3}{2(1)} = -1.5 \)
Now, substitute \( x = -1.5 \) back into the equation to find y:
\( y = (-1.5)^2 + 3(-1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 \)
The vertex is \( (-1.5, -0.25) \)
Key steps:
  • Use \( h = -\frac{b}{2a} \) to find h.
  • Substitute h back into the equation to find y.
Parabolic Curve
The graph of a quadratic equation is a parabolic curve. This curve can open upwards or downwards based on the coefficient of \( x^2 \). In our equation, since the coefficient of \( x^2 \) is positive (a=1), the parabola opens upwards.
To plot the parabolic curve, follow these steps:
  • Identify and plot the y-intercept (0, 2).
  • Find and plot the x-intercepts (-1, 0) and (-2, 0).
  • Determine and plot the vertex (-1.5, -0.25).
Once these key points are plotted, draw a smooth curve passing through them, ensuring it opens upwards.
Remember, the shape of the parabola is symmetrical around the vertex.

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Most popular questions from this chapter

In Exercises \(1-64\), solve each of the given equations. If the equation is quadratic, use the factoring or square root method. If the equation has no real solutions, say so. $$(2 x-3)(x+4)=x(x+9)$$

Suppose that a company's profit \(P,\) in thousands of dollars, on the sale of \(x\) items is given by the equation. $$P=x^{2}-6 x-40 \quad \text { where } x \geq 0$$ (a) Sketch the graph of this equation. Let \(x\) be the horizontal axis and \(P\) be the vertical axis. (b) Use the graph to determine the minimum profit the company earns. (c) How many items does the company sell if it experiences this minimum profit?

In Exercises \(1-64\), solve each of the given equations. If the equation is quadratic, use the factoring or square root method. If the equation has no real solutions, say so. $$(x-6)^{2}=12$$

In Exercises \(65-74\), solve the given equation. For quadratic equations, choose either the factoring method or the square root method, whichever you think is the easier to use. $$(y+3)(y-5)=(2 y+5)(y-3)$$

In Exercises \(85-94\), solve the equations using the square root method. Round off your answers to the nearest hundredth. $$(3 x-0.5)^{2}=10.4$$
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