Algebra 1 : How to graph a quadratic function

Study concepts, example questions & explanations for Algebra 1

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Example Questions

Example Question #2 : Parabolic Functions

What is the minimum possible value of the expression below?

\(\displaystyle 3x^{2}+6x-10\)

Possible Answers:

\(\displaystyle -19\)

The expression has no minimum value.

\(\displaystyle -13\)

\(\displaystyle -7\)

\(\displaystyle -10\)

Correct answer:

\(\displaystyle -13\)

Explanation:

We can determine the lowest possible value of the expression by finding the \(\displaystyle y\)-coordinate of the vertex of the parabola graphed from the equation \(\displaystyle y=3x^{2}+6x-10\). This is done by rewriting the equation in vertex form.

\(\displaystyle 3x^{2}+6x-10\)

\(\displaystyle 3(x^{2}+2x)-10\)

\(\displaystyle 3(x^{2}+2x+1)-3* 1-10\)

\(\displaystyle 3(x+1)^{2}-13\)

The vertex of the parabola \(\displaystyle y=3(x+1)^{2}-13\) is the point \(\displaystyle (-1, -13)\).

The parabola is concave upward (its quadratic coefficient is positive), so \(\displaystyle -13\) represents the minimum value of \(\displaystyle y\). This is our answer.

Example Question #34 : Graphing

What is the vertex of the function \(\displaystyle \small f(x)=2(x-4)^2+7\)? Is it a maximum or minimum?

Possible Answers:

\(\displaystyle \small (4,7)\); maximum

\(\displaystyle \small (-4,7)\); minimum

\(\displaystyle \small (4,7)\); minimum

\(\displaystyle \small (-4,7)\); maximum

Correct answer:

\(\displaystyle \small (4,7)\); minimum

Explanation:

The equation of a parabola can be written in vertex form: \(\displaystyle \small f(x)=a(x-h)^2+k\).

The point \(\displaystyle \small (h,k)\) in this format is the vertex. If \(\displaystyle \small a\) is a postive number the vertex is a minimum, and if \(\displaystyle \small a\) is a negative number the vertex is a maximum.

\(\displaystyle \small f(x)=2(x-4)^2+7\)

In this example, \(\displaystyle \small a=2\). The positive value means the vertex is a minimum.

\(\displaystyle \small h=4\ \text{and}\ k=7\)

\(\displaystyle \small (h,k)=(4,7)\)

Example Question #1 : Graphing Polynomial Functions

Which of the graphs best represents the following function?

\(\displaystyle f(x)=\frac{9}{10}x^2-7x+2\)

Possible Answers:

Graph_line_

Graph_parabola_

Graph_exponential_

Graph_cube_

None of these

Correct answer:

Graph_parabola_

Explanation:

\(\displaystyle f(x)=\frac{9}{10}x^2-7x+2\)

The highest exponent of the variable term is two (\(\displaystyle x^2\)). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

Graph_parabola_

Example Question #1 : Quadratic Functions

What is the equation of a parabola with vertex \(\displaystyle (4,1)\) and \(\displaystyle y\)-intercept \(\displaystyle (0,-7)\)?

Possible Answers:

\(\displaystyle y = -\frac{1}{2} x^{2} -4x - 7\)

\(\displaystyle y = \frac{1}{2} x^{2} -4x - 7\)

\(\displaystyle y = \frac{1}{2} x^{2} -8x - 7\)

\(\displaystyle y = -\frac{1}{2} x^{2} +4x - 7\)

\(\displaystyle y = \frac{1}{2} x^{2} +8x - 7\)

Correct answer:

\(\displaystyle y = -\frac{1}{2} x^{2} +4x - 7\)

Explanation:

From the vertex, we know that the equation of the parabola will take the form \(\displaystyle y = a (x - 4) ^{2}+ 1\) for some \(\displaystyle a\) .

To calculate that \(\displaystyle a\), we plug in the values from the other point we are given, \(\displaystyle x = 0, y=-7\), and solve for \(\displaystyle a\):

\(\displaystyle y = a (x - 4)^{2} + 1\)

\(\displaystyle -7= a \cdot (0 - 4)^{2} + 1\)

\(\displaystyle -7= a \cdot (- 4)^{2} + 1\)

\(\displaystyle -7= 16a + 1\)

\(\displaystyle 16a = -8\)

\(\displaystyle a = -\frac{1}{2}\)

Now the equation is \(\displaystyle y = -\frac{1}{2} (x - 4)^{2} + 1\). This is not an answer choice, so we need to rewrite it in some way.

Expand the squared term:

\(\displaystyle y = -\frac{1}{2} (x^{2} -8x + 16) + 1\)

Distribute the fraction through the parentheses:

\(\displaystyle y = -\frac{1}{2} x^{2} +4x - 8 + 1\)

Combine like terms:

\(\displaystyle y = -\frac{1}{2} x^{2} +4x - 7\)

Example Question #4 : Graphing Polynomial Functions

Possible Answers:

None of the above

\(\displaystyle 3)\) \(\displaystyle y = (x+3)^{2}\)

\(\displaystyle 2)\) \(\displaystyle y = (x-2)^{2}\)

\(\displaystyle 4)\) \(\displaystyle y = -(x-2)^{2} - 2\)

\(\displaystyle 1)\) \(\displaystyle y = x^{2}\)

Correct answer:

\(\displaystyle 3)\) \(\displaystyle y = (x+3)^{2}\)

Explanation:

Starting with \(\displaystyle y = x^{2}\)

\(\displaystyle (x-2)^{2}\) moves the parabola \(\displaystyle x^{2}\) by \(\displaystyle 2\) units to the right.

Similarly \(\displaystyle (x+3)^{2}\) moves the parabola by \(\displaystyle 3\) units to the left.

Hence the correct answer is option \(\displaystyle 3\).

Example Question #281 : Grade 8

Which of the following graphs matches the function \(\displaystyle y=(x-1)^2-2\)?

Possible Answers:

Graph

Graph3

Graph2

Graph4

Graph1

Correct answer:

Graph

Explanation:

Start by visualizing the graph associated with the function \(\displaystyle y=x^2\):

Graph5

Terms within the parentheses associated with the squared x-variable will shift the parabola horizontally, while terms outside of the parentheses will shift the parabola vertically. In the provided equation, 2 is located outside of the parentheses and is subtracted from the terms located within the parentheses; therefore, the parabola in the graph will shift down by 2 units. A simplified graph of \(\displaystyle y=x^2-2\) looks like this:

Graph6

Remember that there is also a term within the parentheses. Within the parentheses, 1 is subtracted from the x-variable; thus, the parabola in the graph will shift to the right by 1 unit. As a result, the following graph matches the given function \(\displaystyle y=(x-1)^2-2\) :

Graph

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