How To Find The Standard Form Of A Graph

Quadratic Functions

Contents: This page corresponds to § 3.1 (p. 244) of the text.

Graphs

A quadratic function is one of the form f(x) = ax² + bx + c, where a, b, and c are numbers with a non equal to zero.

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

All parabolas are symmetric with respect to a line chosen 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 decide a line. This means that if you are given any two points in the plane, then there is i and only i line that contains both points. A similar statement can be fabricated most points and quadratic functions.

Given three points in the plane that have unlike offset coordinates and do non lie on a line, there is exactly 1 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 respective function is shown in the text box below the graph. If yous 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. (Run into the section on manipulating graphs.)

Example ane.

Sketch the graph of y = x²/2. Starting with the graph of y = tenⁱⁱ, we compress by a gene of one half. This ways that for each point on the graph of y = xⁱⁱ, we depict a new point that is ane half of the mode from the x-centrality to that indicate.

Example 2.

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

Practise one :

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

(b) Sketch the graph of y = -(x - 5)^two + three. Reply

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Standard Course

The functions in parts (a) and (b) of Exercise 1 are examples of quadratic functions in standard class. When a quadratic role is in standard form, then information technology is piece of cake 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 class. 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 class by completing the square. (See the department on solving equations algebraically to review completing the foursquare.) The steps that we utilize in this section for completing the square will wait a little different, because our chief goal here is not solving an equation.

Notation that when a quadratic part is in standard form it is likewise easy to find its zeros by the square root principle.

Instance iii.

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

f(ten) = x² - 6x + 7.

= (x² - 6x )+ seven. Group the 10² and x terms and then complete the foursquare on these terms.

= (x^two - 6x + 9 - 9) + 7.

We need to add ix considering it is the square of one half the coefficient of x, (-6/ii)² = ix. When we were solving an equation we simply added nine to both sides of the equation. In this setting we add and decrease 9 and so that we exercise not modify the office.

= (10ⁱⁱ - 6x + ix) - 9 + 7. Nosotros see that 10² - 6x + ix is a perfect square, namely (x - 3)².

f(x) = (10 - three)² - 2. This is standard course.

From this result, 1 easily finds the vertex of the graph of f is (3, -ii).

To detect the zeros of f, we set f equal to 0 and solve for 10.

(ten - 3)² - two = 0.

(x - 3)ⁱⁱ = ii.

(x - three) = ± sqrt(2).

10 = three ± sqrt(2).

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

If the coefficient of 10² is not one, then we must factor this coefficient from the tenⁱⁱ and 10 terms before proceeding.

Case 4.

Write f(ten) = -2x² + 2x + iii in standard form and discover the vertex of the graph of f.

f(ten) = -2x² + 2x + 3.

= (-2x² + 2x) + iii.

= -2(x^two - x) + 3.

= -2(10² - 10 + one/four - 1/4) + three.

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

= -2(x² - x + 1/4) -2(-1/4) + three.

Note that everything in the parentheses is multiplied by -2, so when we remove -ane/iv from the parentheses, nosotros must multiply it past -2.

= -two(x - i/2)ⁱⁱ + ane/two + iii.

= -2(10 - 1/2)² + 7/2.

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

Practice 2 :

Write f(10) = 3x^two + 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 non the easiest style to detect the vertex of a parabola. If the graph of a quadratic part has 2 x-intercepts, then the line of symmetry is the vertical line through the midpoint of the x-intercepts.

The x-intercepts of the graph above are at -five and 3. The line of symmetry goes through -1, which is the average of -5 and 3. (-5 + three)/two = -2/two = -1. In one case nosotros know that the line of symmetry is x = -1, then we know the first coordinate of the vertex is -one. The 2nd coordinate of the vertex can be found by evaluating the function at ten = -1.

Example 5.

Find the vertex of the graph of f(ten) = (x + 9)(x - v).

Since the formula for f is factored, information technology is like shooting fish in a barrel to find the zeros: -9 and 5.

The boilerplate of the zeros is (-9 + 5)/ii = -iv/2 = -2. So, the line of symmetry is x = -2 and the first coordinate of the vertex is -2.

The second coordinate of the vertex is f(-2) = (-ii + 9)(-2 - v) = 7*(-seven) = -49.

Therefore, the vertex of the graph of f is (-two, -49).

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Applications

Example 6.

A rancher has 600 meters of fence to enclose a rectangular corral with another debate dividing it in the middle equally in the diagram beneath.

Every bit indicated in the diagram, the four horizontal sections of fence volition each be x meters long and the three vertical sections will each be y meters long.

The rancher's goal is to use all of the fence and enclose the largest possible surface area.

The two rectangles each accept area xy, so we take

total expanse: A = 2xy.

There is non much nosotros can do with the quantity A while it is expressed every bit a production of two variables. However, the fact that we take only 1200 meters of contend available leads to an equation that ten and y must satisfy.

3y + 4x = 1200.

3y = 1200 - 4x.

y = 400 - 4x/3.

We now accept y expressed every bit a part of x, and we can substitute this expression for y in the formula for total area A.

A = 2xy = 2x (400 -4x/3).

We need to observe the value of ten that makes A as large every bit possible. A is a quadratic function of x, and the graph opens downwards, then the highest indicate on the graph of A is the vertex. Since A is factored, the easiest fashion to find the vertex is to detect the x-intercepts and average.

2x (400 -4x/3) = 0.

2x = 0 or 400 -4x/3 = 0.

x = 0 or 400 = 4x/3.

ten = 0 or 1200 = 4x.

x = 0 or 300 = x.

Therefore, the line of symmetry of the graph of A is 10 = 150, the average of 0 and 300.

At present that we know the value of x respective to the largest area, we can notice the value of y by going back to the equation relating 10 and y.

y = 400 - 4x/3 = 400 -4(150)/3 = 200.