Question

1. tell if the following statements are true or false regarding the graph of the function $f(x) = \frac{1}{2}(x + 5)^2 - 8$?

a. the domain is $x \geq -5$

b. the range is $y \leq -8$

c. it opens upward

d. the vertex is at $(-5, -8)$

e. it has a minimum value of $-8$

f. the $y$-intercept is $(0, -8)$

g. its axis of symmetry is $x = 5$

h. the graph is reflected

Explanation:

Step1: Recall the properties of quadratic functions in vertex form \( f(x) = a(x - h)^2 + k \)

• The vertex is \((h, k)\), the axis of symmetry is \(x = h\), the direction it opens depends on the sign of \(a\) (upward if \(a>0\), downward if \(a<0\)), the domain of a quadratic function is all real numbers, the range is \(y\geq k\) if \(a>0\) and \(y\leq k\) if \(a<0\), the y - intercept is found by setting \(x = 0\) and solving for \(y\).

Given the function \(f(x)=\frac{1}{2}(x + 5)^2-8=\frac{1}{2}(x-(-5))^2+(-8)\), so \(a=\frac{1}{2}\), \(h=-5\), \(k = - 8\)

Step2: Analyze each statement

Statement A: The domain is \(x\geq - 5\)

The domain of a quadratic function (a polynomial function of degree 2) is all real numbers (\((-\infty,\infty)\)) because there are no restrictions on the values of \(x\) that we can plug into the function. So the statement is False.

Statement B: The range is \(y\leq - 8\)

Since \(a=\frac{1}{2}>0\), the parabola opens upward. For a parabola in vertex form \(y=a(x - h)^2 + k\) with \(a>0\), the range is \(y\geq k\). Here \(k=-8\), so the range is \(y\geq - 8\), not \(y\leq - 8\). The statement is False.

Statement C: It opens upward

Since \(a=\frac{1}{2}>0\), when the coefficient \(a\) of \((x - h)^2\) is positive, the parabola opens upward. So the statement is True.

Statement D: The vertex is at \((-5,-8)\)

For the function \(y=a(x - h)^2 + k\), the vertex is \((h,k)\). In our function \(f(x)=\frac{1}{2}(x+5)^2-8=\frac{1}{2}(x-(-5))^2+(-8)\), \(h=-5\) and \(k = - 8\), so the vertex is \((-5,-8)\). The statement is True.

Statement E: It has a minimum value of \(-8\)

Since the parabola opens upward (\(a=\frac{1}{2}>0\)), the vertex is the minimum point of the parabola. The y - coordinate of the vertex is \(k=-8\), so the minimum value of the function is \(-8\). The statement is True.

Statement F: The y - intercept is \((0,-8)\)

To find the y - intercept, set \(x = 0\) in the function:

\(f(0)=\frac{1}{2}(0 + 5)^2-8=\frac{1}{2}\times25-8=\frac{25}{2}-8=\frac{25 - 16}{2}=\frac{9}{2}=4.5\)

So the y - intercept is \((0,\frac{9}{2})\) or \((0,4.5)\), not \((0,-8)\). The statement is False.

Statement G: Its axis of symmetry is \(x = 5\)

For a parabola in vertex form \(y=a(x - h)^2 + k\), the axis of symmetry is \(x = h\). Here \(h=-5\), so the axis of symmetry is \(x=-5\), not \(x = 5\). The statement is False.

Statement H: The graph is reflected

A reflection of a parabola \(y=(x - h)^2 + k\) would occur if \(a<0\) (reflection over the x - axis) or if there is a reflection over the y - axis (change in the sign of \(h\)). Since \(a=\frac{1}{2}>0\) and there is no reflection over the y - axis (the original form of the parabola \(y=(x + 5)^2\) is a horizontal shift of \(y=x^2\) to the left by 5 units, and then vertical stretch and shift), the graph is not reflected (in the sense of a reflection that would change the direction of opening or the basic shape in a reflective way other than the normal transformation). The statement is False.

Answer:

A. False
B. False
C. True
D. True
E. True
F. False
G. False
H. False