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Chapter 3: Problem 36

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. $$f(x)=3 x^{2}-2 x-4$$

Short Answer

Expert verified
The vertex of the parabola is \((1/3, -13/3)\). The x-intercepts are \((2/3, 0)\) and \((-2, 0)\), and the y-intercept is \((0, -4)\). The equation of the axis of symmetry is \(x = 1/3\). The function's domain and range are \(-鈭� < x < 鈭瀄) and \(-13/3 鈮� y < 鈭瀄) respectively.

Step by step solution

01

Find the Vertex

The vertex form of a quadratic function is \(f(x)=a(x-h)^{2}+k\), where \((h, k)\) is the vertex. The x-coordinate of the vertex, \(h\), is given by \(-b/2a\). Substituting \(a = 3\) and \(b = -2\) into this formula, we get \(h = -(-2)/(2*3) = 1/3\). We can then substitute \(h = 1/3\) back into the equation to get the y-coordinate \(k\) of the vertex. So, \(k = 3*(1/3)^2 - 2*(1/3) -4 = -13/3\). So, the vertex is \((1/3, -13/3)\).
02

Find the Intercepts

The x-intercept(s) of a function are obtained by setting \(f(x) = 0\). So, we solve \(3x^{2}-2x-4 = 0\). This is a quadratic equation and can be solved by using the quadratic formula \(x = [-b 卤 sqrt(b^{2}-4ac)]/(2a)\). Solving the equation we find \(x = 2/3\) and \(x = -2\). The y-intercept is obtained by setting \(x = 0\), so we substitute \(x = 0\) into \(f(x)\) and get \(f(0) = -4\). So, we have two x-intercepts \((2/3, 0)\) and \((-2, 0)\) and one y-intercept \((0, -4)\).
03

Sketch the Graph

To sketch the graph of the function, we plot the vertex, \((1/3, -13/3)\), the x-intercepts, \((2/3, 0)\) and \((-2, 0)\), and the y-intercept, \((0, -4)\). Because the coefficient of \(x^{2}\) is positive, the parabola opens upwards. We then sketch a U-shaped curve that goes through these points.
04

Find the Axis of Symmetry

The axis of symmetry of a parabola is the vertical line \(x = h\), where \(h\) is the x-coordinate of the vertex. So, in this case, the axis of symmetry is \(x = 1/3\).
05

Determine the Domain and Range

The domain of a quadratic function is all real numbers because the function is defined for all real values of \(x\). So, the domain is \(-鈭� < x < 鈭瀄). The range of a quadratic function that opens upwards and has a vertex \((h, k)\) is all numbers greater than or equal to the y-coordinate of the vertex. So, the range in this case is \(-13/3 鈮� y < 鈭瀄)

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