Question

solve the quadratic by graphing

1. ((x - 2)^2 - 3 = 1)

solutions are:
x = _________ and x = _____________

1. (-(x + 4)^2 + 5 = -4)

solutions are:
x = _________ and x = _____________
find the vertex and the solutions given the graph. write the quadratic function in vertex form, standard

Explanation:

Problem 13: \((x - 2)^2 - 3 = 1\)

Step 1: Rewrite the equation

Rewrite the equation as \((x - 2)^2 - 4 = 0\). The quadratic function is \(y=(x - 2)^2 - 4\), which is a parabola in vertex form \(y = a(x - h)^2 + k\) with vertex \((h,k)=(2,-4)\) and \(a = 1\) (opens upward).

Step 2: Find the x - intercepts

To find the solutions (x - intercepts), we set \(y = 0\):
\[

$$\begin{align*} (x - 2)^2 - 4&=0\\ (x - 2)^2&=4\\ x - 2&=\pm2 \end{align*}$$

\]

Step 3: Solve for x

Case 1: \(x - 2 = 2\), then \(x=2 + 2=4\)
Case 2: \(x - 2=-2\), then \(x=2-2 = 0\)

Step 1: Rewrite the equation

Rewrite the equation as \(-(x + 4)^2+9 = 0\) or \((x + 4)^2=9\). The quadratic function is \(y=-(x + 4)^2 + 9\), which is a parabola in vertex form \(y=a(x - h)^2 + k\) with vertex \((h,k)=(-4,9)\) and \(a=- 1\) (opens downward).

Step 2: Find the x - intercepts

Set \(y = 0\):
\[

$$\begin{align*} -(x + 4)^2+9&=0\\ (x + 4)^2&=9\\ x + 4&=\pm3 \end{align*}$$

\]

Step 3: Solve for x

Case 1: \(x+4 = 3\), then \(x=3 - 4=-1\)
Case 2: \(x + 4=-3\), then \(x=-3-4=-7\)

Answer:

\(x = 0\) and \(x = 4\)

Problem 14: \(-(x + 4)^2 + 5=-4\)