Question

The quadratic function f(x) = x2 + 2x – 1 is expressed in standard form as
a. f(x) = (x + 1)2 + 1
b. f(x) = (x + 1)2 – 2
d. f(x) = (x + 1)2 - 1
c. f(x) = (x + 1)2 + 2
​

Answer 1

Answer:

Step-by-step explanation:

Graphs

A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are numbers with a not equal to zero.

The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downward and vary in "width" or "steepness", but they all have the same basic "U" shape. The picture below shows three graphs, and they are all parabolas.

All parabolas are symmetric with respect to a line called the axis of symmetry. A parabola intersects its axis of symmetry at a point called the vertex of the parabola.

You know that two points determine a line. This means that if you are given any two points in the plane, then there is one and only one line that contains both points. A similar statement can be made about points and quadratic functions.

Given three points in the plane that have different first coordinates and do not lie on a line, there is exactly one quadratic function f whose graph contains all three points. The applet below illustrates this fact. The graph contains three points and a parabola that goes through all three. The corresponding function is shown in the text box below the graph. If you drag any of the points, then the function and parabola are updated.

Many quadratic functions can be graphed easily by hand using the techniques of stretching/shrinking and shifting (translation) the parabola y = x2 . (See the section on manipulating graphs.)

Example 1.

Sketch the graph of y = x2/2. Starting with the graph of y = x2, we shrink by a factor of one half. This means that for each point on the graph of y = x2, we draw a new point that is one half of the way from the x-axis to that point.

Example 2.

Sketch the graph of y = (x - 4)^2 - 5. We start with the graph of y = x2 , shift 4 units right, then 5 units down.

Exercise 1:

(a) Sketch the graph of y = (x + 2)2 - 3. Answer

(b) Sketch the graph of y = -(x - 5)2 + 3. Answer

Return to Contents

Standard Form

The functions in parts (a) and (b) of Exercise 1 are examples of quadratic functions in standard form. When a quadratic function is in standard form, then it is easy to sketch its graph by reflecting, shifting, and stretching/shrinking the parabola y = x2.

The quadratic function f(x) = a(x - h)2 + k, a not equal to zero, is said to be in standard form. If a is positive, the graph opens upward, and if a is negative, then it opens downward. The line of symmetry is the vertical line x = h, and the vertex is the point (h,k).

Any quadratic function can be rewritten in standard form by completing the square. (See the section on solving equations algebraically to review completing the square.) The steps that we use in this section for completing the square will look a little different, because our chief goal here is not solving an equation.

Note that when a quadratic function is in standard form it is also easy to find its zeros by the square root principle.

Example 3.

Write the function f(x) = x2 - 6x + 7 in standard form. Sketch the graph of f and find its zeros and vertex.

f(x) = x2 - 6x + 7.

= (x2 - 6x )+ 7.        Group the x2 and x terms and then complete the square on these terms.

= (x2 - 6x + 9 - 9) + 7.

We need to add 9 because it is the square of one half the coefficient of x, (-6/2)2 = 9. When we were solving an equation we simply added 9 to both sides of the equation. In this setting we add and subtract 9 so that we do not change the function.

= (x2 - 6x + 9) - 9 + 7. We see that x2 - 6x + 9 is a perfect square, namely (x - 3)2.

f(x) = (x - 3)2 - 2. This is standard form.

From this result, one easily finds the vertex of the graph of f is (3, -2).

To find the zeros of f, we set f equal to 0 and solve for x.

(x - 3)2 - 2 = 0.

(x - 3)2 = 2.

(x - 3) = ± sqrt(2).

x = 3 ± sqrt(2).

To sketch the graph of f we shift the graph of y = x2 three units to the right and two units down.

If the coefficient of x2 is not 1, then we must factor this coefficient from the x2 and x terms before proceeding.

Example 4.

Write f(x) = -2x2 + 2x + 3 in standard form and find the vertex of the graph of f.

f(x) = -2x2 + 2x + 3.

= (-2x2 + 2x) + 3.

= -2(x2 - x) + 3.

= -2(x2 - x + 1/4 - 1/4) + 3.

We add and subtract 1/4, because (-1/2)2 = 1/4, and -1 is the coefficient of x.

= -2(x2 - x + 1/4) -2(-1/4) + 3.

Note that everything in the parentheses is multiplied by -2, so when we remove -1/4 from the parentheses, we must multiply it by -2.

= -2(x - 1/2)2 + 1/2 + 3.

= -2(x - 1/2)2 + 7/2.

The vertex is the point (1/2, 7/2). Since the graph opens downward (-2 < 0), the vertex is the highest point on the graph.

Exercise 2:

Write f(x) = 3x2 + 12x + 8 in standard form. Sketch the graph of f ,find its vertex, and find the zeros of f. Answer

Alternate method of finding the vertex

In some cases completing the square is not the easiest way to find the vertex of a parabola. If the graph of a quadratic function has two x-intercepts, then the line of symmetry is the vertical line through the midpoint of the x-intercepts.

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

The quadratic function f(x) = x² + 2x – 1 is expressed in standard form as

$\sf a. \: \: f(x) = {(x + 1)}^{2} + 1$

$\sf b. \: \: f(x) = {(x + 1)}^{2} - 2$

$\sf c. \: \: f(x) = {(x + 1)}^{2} - 1$

$\sf d. \: \: f(x) = {(x + 1)}^{2} + 2$

EVALUATION

Here the given Quadratic equation is

$\sf f(x) = {x}^{2} + 2x - 1$

Which can be rewritten as below

$\sf f(x) = {x}^{2} + 2x - 1$

$\sf \implies f(x) = {x}^{2} + 2x + 1 - 2$

$\sf \implies f(x) = {x}^{2} + 2.x .1 + {(1)}^{2} - 2$

$\sf \implies f(x) = {(x + 1)}^{2} - 2$

FINAL ANSWER

Hence the correct option is

$\sf b. \: \: f(x) = {(x + 1)}^{2} - 2$

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