Are these ordered pairs in direct variation (1 2) (2 3) (3 4)
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A discontinuity is a point that cannot exist because the x-coordinate would cause a problem in the equation. If you have a polynomial in the denominator, you must find which values of x would cause the polynomial in the denominator to evaluate to zero. Since division by zero is undefined, that would cause a discontinuity.
Let's look at your function.
is in the numerator. It is defined for every value of x. There is no problem there.
is in the denominator. This is a function defined for every value of x, but since it is in the denominator, we must exclude the x-value that would cause this polynomial to evaluate to zero.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
Let's look at your function.
We set it equal to zero and solve the equation for x.
For x = 0, the denominator has a value of zero, so at this point there is a discontinuity in function f(x).
The answer is:
1. Determine y-intercept.
<em>The y-intercept is (0,-3). Substitute x = 0 in the equation as we get f(x) = -3 as the y-intercept.</em>
2. Determine the zeros.
<em>Factor the polynomials first. </em> <em>This is the factored form. The zeros are the roots of equation. Therefore, the zeros are 1/2 and -3/2</em>
3. Determine the Axis of Symmetry
<em>We can solve this by using the formula of </em> <em>However, I'll be solving the Axis of Symmetry with Calculus instead.</em>
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<em>Then let f'(x) = 0 to find the Axis of Symmetry.</em>
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<em>Therefore, the Axis of Symmetry is -1/2</em>
4. Determine the vertex.
<em>Substitute the value of Axis of Symmetry in f(x).</em>
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<em>Therefore the vertex is at (-1/2, -4)</em>
5. Does the vertex representing max-pont or min-point?
<em>The vertex represents the minimum point. The graph is upward, meaning the minimum point is the point that gives the LOWEST Y-VALUE. </em>
7. Graph f(x)
<em>Unfortunately I won't be able to graph. But I can tell you to graph a parabola that has the most curve at the vertex and intercepts y-axis at (0,-3).</em>
<em>--------------------------- </em>End of Part 1 --------------------------------
2. Write the general form of Quadratic Function in standard form.
<em>The graph can be easily determined about the y-intercept and being an upward or downward parabola along with how narrow or wide the parabola is. </em>
<em>For example, c value is the y-intercept as defined. When a > 0, the parabola is upward and when a < 0, the parabola is downward. The more value of | a | is, the more narrow it will be and the less value of | a | it is, the wider it will be.</em>
3. Write in Factored Form
<em>These are factored forms with different types of operators.</em>
<em>The equation can be easily determined about the roots of equation. For example, if the function is in f(x) = (x+2)(x-1) Then the roots would be x = -2 and 1 as the graph will intercept x-axis at (-2,0) and (1,0)</em>
4. Write in Vertex Form.
<em>The equation can be easily determined for the vertex, axis of symmetry and the same narrow/wide/upward/downward parabola again.</em>
<em>The vertex is at (h,k) and the axis of symmetry is at x = h.</em>
<em>For example, </em><em />
<em>The vertex would be at (2,3) and the axis of symmetry is x = 2.</em>