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Chapter 3: Problem 79

A parabola that opens upward has its vertex at \((1,2)\) Describe as much as you can about the parabola based on this information. Include in your discussion the number of \(x\) -intercepts (if any) for the parabola.

Short Answer

Expert verified
The given parabola's equation should be \(y = a(x - 1)^2 + 2\), as per the vertex form for parabolas. As long as 'a' is not equal to 0, the parabola will have x-intercepts. However, without knowing the exact value of 'a', it is impossible to specify the exact x-intercepts. Finally, since 'a' is positive according to the problem's context, the parabola opens upwards.

Step by step solution

01

Understanding Vertex Form of Parabola

The vertex form of a parabola is \(y = a(x - h)^2 + k\) where \((h, k)\) is the vertex of the parabola. As per the problem, the vertex of the parabola is given as \((1,2)\). Therefore, the equation of the parabola is \(y = a(x - 1)^2 + 2\). Note that 'a' here determines the direction and width of the parabola. If 'a' is greater than 0, parabola opens upward and if 'a' is less than 0, parabola opens downward. As per the problem, parabola opens upward so we assume 'a' to be positive without knowing its exact value.
02

Finding X-intercepts of Parabola

To find the point where the parabola intercepts x-axis, we set the 'y' in the equation of the parabola to zero and solve for 'x'. We get \(0 = a(x - 1)^2 + 2\). Solving this equation can give the x-intercepts (if they exist). However, without a numerical value for 'a', we can not definitively compute x-intercepts. We can, however, say that it will have x-intercepts as long as 'a' is not equal to 0. The x-intercepts could be real or complex numbers depending on the value of 'a'.

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