Question

6 multiple choice 1 point which is an equation of the parabola graphed in the accompanying diagram? y 5 4 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 x -1 -2 -3 -4 -5 y = x² - 4 y = x² + 4 y = -x² + 4 y = -x² - 4 7 multiple choice 1 point the graph of 2ˣ = y has a point on the: positive portion of the x - axis negative portion of the x - axis negative portion of the y - axis positive portion of the y - axis

Explanation:

Step1: Identify vertex and direction of parabola

The parabola opens downwards (since it has a maximum - like a frown), so the coefficient of $x^{2}$ must be negative. The vertex of the parabola is at $(0,4)$. The general form of a parabola is $y = a(x - h)^{2}+k$, where $(h,k)$ is the vertex. Here $h = 0,k = 4$ and $a<0$.

Step2: Analyze the options

• For $y=x^{2}-4$, the coefficient of $x^{2}$ is positive and the vertex is $(0, - 4)$.
• For $y=x^{2}+4$, the coefficient of $x^{2}$ is positive and the vertex is $(0,4)$.
• For $y=-x^{2}+4$, the coefficient of $x^{2}$ is negative and the vertex is $(0,4)$.
• For $y=-x^{2}-4$, the coefficient of $x^{2}$ is negative and the vertex is $(0,-4)$.

for second question:

Step1: Consider the properties of the exponential function $y = 2^{x}$

The exponential function $y = a^{x}$ where $a>1$ (here $a = 2$) has the following properties: when $x = 0$, $y=2^{0}=1$; when $x>0$, $y>1$; when $x<0$, $0

Step2: Analyze the position of points

• If $x$ is on the positive portion of the $x -$axis ($x>0$), then $y = 2^{x}>1$.
• If $x$ is on the negative portion of the $x -$axis ($x<0$), then $0
• The function $y = 2^{x}$ never intersects the $x -$axis because $2^{x}>0$ for all real - valued $x$. When $x = 0$, $y = 1$, so it has a point $(0,1)$ which is on the positive portion of the $y -$axis.

Answer:

$y=-x^{2}+4$