Problem 20 (a) find the vertex and the axis... [FREE SOLUTION] (2024)

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Chapter 11: Problem 20

(a) find the vertex and the axis of symmetry and (b) graph the function. $$ f(x)=x^{2}+2 x-5 $$

Short Answer

Expert verified

The vertex is \( (-1, -6) \) and the axis of symmetry is \( x = -1 \).

Step by step solution

01

Rewrite the Function in Vertex Form

To find the vertex, first, rewrite the quadratic function in vertex form: \[ f(x) = a(x - h)^2 + k \] where \( h \) is the x-coordinate of the vertex and \( k \) is the y-coordinate of the vertex. First, complete the square for the function \( f(x) = x^2 + 2x - 5 \).

02

Complete the Square

To complete the square, take the coefficient of \( x \), divide it by 2, and square it: \( (2 / 2)^2 = 1 \). Add and subtract this square inside the function: \[ f(x) = x^2 + 2x + 1 - 1 - 5 = (x + 1)^2 - 6 \].

03

Identify the Vertex

Now the equation is in vertex form \( f(x) = (x + 1)^2 - 6 \). The vertex \( (h, k) \) is \( (-1, -6) \).

04

Determine the Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. For the equation \( f(x) = (x + 1)^2 - 6 \), it is \( x = -1 \).

05

Graph the Function

To graph \( f(x) = x^2 + 2x - 5 \): 1. Plot the vertex point \( (-1, -6) \). 2. Draw the axis of symmetry \( x = -1 \). 3. Plot additional points by choosing values for \( x \) and solving for \( f(x) \), such as \( x = -2 \) and \( x = 0 \). Draw the parabola opening upwards.

Key Concepts

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

quadratic function

A quadratic function is a type of polynomial where the highest exponent of the variable (usually x) is 2. The general form is f(x) = ax^2 + bx + c where a, b, and c are constants. Quadratic functions help us model many real-world situations, like projectile motion. One key feature is the parabolic shape of the graph. The peak or the lowest point of this parabola, called the vertex, plays a crucial role in understanding the function’s properties. Knowing how to find the vertex and graph the parabola accurately is important for solving many mathematical problems.

vertex form

The vertex form of a quadratic function is a convenient way to identify the vertex and the graph's shape. It’s given by: f(x) = a(x - h)^2 + k where - (h, k) is the vertex of the parabola - a determines the direction (opening up or down) and the width of the parabola. To convert a quadratic function from its standard form to vertex form, you often need to complete the square. This conversion makes it easier to graph the function and analyze its key features. For example, the quadratic function f(x) = x^2 + 2x - 5 can be rewritten in vertex form by completing the square, resulting in f(x) = (x + 1)^2 - 6. From this, we can easily identify that the vertex is at (h, k) = (-1, -6).

graphing parabolas

Graphing parabolas involves several steps: - Identify the vertex, which is the point (h, k). This is the lowest or highest point on the graph. - Determine the axis of symmetry, a vertical line through the vertex, given by x = h. - Calculate additional points on either side of the axis of symmetry for accuracy. - Connect these points smoothly to draw a symmetric curve. For example, graphing the function f(x) = x^2 + 2x - 5 involves: 1. Finding the vertex at - ( - 1, - 6) - 2. Drawing the axis of symmetry at - x = -1 - 3. Selecting other x-values and calculating corresponding f(x) values to plot additional points. - Once we have all necessary points, we draw the parabola opening upwards, since a > 0.

completing the square

Completing the square is a technique used to rewrite a quadratic function in vertex form. Here are the steps: - Start with the general form f(x) = ax^2 + bx + c - Focus on the x-terms (ignore c initially). - Add and subtract a term that makes a perfect square trinomial. - Factor the trinomial into a square of a binomial. - Combine constants outside the trinomial. Let’s apply this to f(x) = x^2 + 2x - 5. - We look at x^2 + 2x. - Half of 2 (the coefficient of x) and square it: (2 / 2)^2 = 1. - Add and subtract this inside the function: x^2 + 2x + 1 - 1 - 5 - Group the perfect square trinomial: ( x + 1)^2 - 6 - This gives us the vertex form: (x + 1)^2 - 6. Completing the square helps us to understand the quadratic function deeply, making it easier to solve and graph.

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Problem 20 (a) find the vertex and the axis... [FREE SOLUTION] (3)

Most popular questions from this chapter

Solve. $$ (x+1)^{2}=-9 $$Write an equation for a function having a graph with the same shape as thegraph of \(f(x)=\frac{3}{5} x^{2},\) but with the given point as the vertex. $$ (-4,-2) $$Complete the square to find the \(x\) -intercepts of each function given by theequation listed. $$ g(x)=x^{2}+5 x+2 $$A quadratic function has \((4,0)\) as one of its intercepts and \((-1,7)\) as itsvertex. Find an equation for the function.If \(f(x)=\left(x-\frac{1}{3}\right)\left(x^{2}+6\right)\) and \(g(x)=\)\(\left(x-\frac{1}{3}\right)\left(x^{2}-\frac{2}{3}\right),\) find all \(a\) forwhich \((f+g)(a)=0\).
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Problem 20 (a) find the vertex and the axis... [FREE SOLUTION] (2024)
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