Q. What is True of the Function \( g(x) = -2x^2 + 5 \)?

Solution

1. Identify the type of function.

   The function is a quadratic polynomial because it has the form \(g(x)=ax^2+bx+c\). Here \(g(x)=-2x^2+5\) with \(a=-2\), \(b=0\), \(c=5\).

2. Determine the direction the parabola opens.

   Because the leading coefficient \(a=-2\) is negative, the parabola opens downward. Therefore the graph has a maximum value and no minimum.

3. Find the vertex (the maximum point).

   The x-coordinate of the vertex of \(ax^2+bx+c\) is \(-\dfrac{b}{2a}\). Here \(b=0\), so

   \(x_{\text{vertex}}=-\dfrac{0}{2(-2)}=0\).

   The y-coordinate is \(g(0)\):

   \(g(0)=-2(0)^2+5=5\).

   Thus the vertex is \((0,5)\). This is the maximum point.

4. Axis of symmetry.

   The axis of symmetry is the vertical line through the vertex, so it is \(x=0\).

5. Domain.

   As a polynomial, the domain is all real numbers: \((-\infty,\infty)\).

6. Range.

   Because the parabola opens downward and the maximum y-value is 5 at the vertex, the range is \((-\infty,5]\).

7. y-intercept.

   Set \(x=0\): \(g(0)=5\). The y-intercept is the point \((0,5)\).

8. x-intercepts (zeros).

   Solve \( -2x^2+5=0\). Divide both sides by \(-2\):

   \(x^2=\dfrac{5}{2}\).

   Take square roots:

   \(x=\pm\sqrt{\dfrac{5}{2}}=\pm\dfrac{\sqrt{10}}{2}\).

   The x-intercepts are \(\left(-\dfrac{\sqrt{10}}{2},0\right)\) and \(\left(\dfrac{\sqrt{10}}{2},0\right)\).

9. Symmetry.

   Compute \(g(-x)=-2(-x)^2+5=-2x^2+5=g(x)\). The function is even, so its graph is symmetric about the y-axis.

10. Increasing and decreasing intervals.

    Differentiate: \(g'(x)=-4x\).

    Set \(g'(x)>0\) to find where the function increases: \(-4x>0\) gives \(x<0\). So \(g\) is increasing on \((-\infty,0)\).

    Set \(g'(x)<0\) to find where it decreases: \(-4x<0\) gives \(x>0\). So \(g\) is decreasing on \((0,\infty)\).

11. Concavity.

    Second derivative \(g”(x)=-4\), which is negative for all x, so the graph is concave down everywhere.