Question

answer the following questions about the given quadratic equation.
y = x² + 16
a) does the graph open up or down?
up
down
b) what is the vertex of the graph? enter your answer as an ordered pair.
c) identify the x-intercept(s). enter your answer as an ordered pair. use a comma to separate answers as needed. if there are none, enter none.

Explanation:

Part (a)

Step1: Recall quadratic form

The standard form of a quadratic equation is \( y = ax^2 + bx + c \). For \( y = x^2 + 16 \), \( a = 1 \), \( b = 0 \), \( c = 16 \).

Step2: Determine direction

If \( a>0 \), the parabola opens up; if \( a<0 \), it opens down. Here, \( a = 1>0 \), so it opens up.

Step1: Recall vertex formula

For \( y = ax^2 + bx + c \), the x - coordinate of the vertex is \( x=-\frac{b}{2a} \). For \( y = x^2 + 16 \), \( a = 1 \), \( b = 0 \).

Step2: Calculate x - coordinate

\( x=-\frac{0}{2\times1}=0 \).

Step3: Find y - coordinate

Substitute \( x = 0 \) into \( y=x^2 + 16 \), we get \( y=0^2 + 16=16 \). So the vertex is \( (0,16) \).

Step1: Recall x - intercept definition

To find x - intercepts, set \( y = 0 \) and solve for \( x \). So we solve \( 0=x^2 + 16 \).

Step2: Solve the equation

\( x^2=- 16 \). Since the square of a real number cannot be negative, there are no real solutions for \( x \). So there are no x - intercepts.

Answer:

Up

Part (b)