Answer :
To find the [tex]\( x \)[/tex]-intercepts of a parabola with vertex [tex]\((-4, 2)\)[/tex] and [tex]\( y \)[/tex]-intercept [tex]\((0, -30)\)[/tex], let's proceed step by step.
### Step 1: Determine the Equation of the Parabola
We start with the vertex form of a quadratic equation:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\( (h, k) \)[/tex] is the vertex. Here, [tex]\( h = -4 \)[/tex] and [tex]\( k = 2 \)[/tex]. Substituting these values, we get:
[tex]\[ y = a(x + 4)^2 + 2 \][/tex]
### Step 2: Use the y-intercept to Find [tex]\( a \)[/tex]
We are given the [tex]\( y \)[/tex]-intercept [tex]\((0, -30)\)[/tex]. This point is on the parabola, so we can substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = -30 \)[/tex] into the equation to solve for [tex]\( a \)[/tex]:
[tex]\[ -30 = a(0 + 4)^2 + 2 \][/tex]
[tex]\[ -30 = a(16) + 2 \][/tex]
[tex]\[ -30 - 2 = 16a \][/tex]
[tex]\[ -32 = 16a \][/tex]
[tex]\[ a = -2 \][/tex]
### Step 3: Write the Complete Equation
Now that we know [tex]\( a = -2 \)[/tex], we can write the full equation of the parabola:
[tex]\[ y = -2(x + 4)^2 + 2 \][/tex]
### Step 4: Find the [tex]\( x \)[/tex]-intercepts
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]. Setting [tex]\( y = 0 \)[/tex] in the equation and solving for [tex]\( x \)[/tex] we get:
[tex]\[ 0 = -2(x + 4)^2 + 2 \][/tex]
[tex]\[ -2(x + 4)^2 + 2 = 0 \][/tex]
[tex]\[ -2(x + 4)^2 = -2 \][/tex]
[tex]\[ (x + 4)^2 = 1 \][/tex]
Taking the square root of both sides:
[tex]\[ x + 4 = 1 \quad \text{or} \quad x + 4 = -1 \][/tex]
[tex]\[ x = 1 - 4 = -3 \quad \text{or} \quad x = -1 - 4 = -5 \][/tex]
### Step 5: Verify the Intercepts
The solutions to the equation are [tex]\( x = -3 \)[/tex] and [tex]\( x = -5 \)[/tex].
Thus, the [tex]\( x \)[/tex]-intercepts of the parabola are:
[tex]\[ (-5, 0) \quad \text{and} \quad (-3, 0) \][/tex]
### Final Answer
[tex]\[ \left(x_1, y_1\right), \left(x_2, y_2\right) = (-5, 0), (-3, 0) \][/tex]
### Step 1: Determine the Equation of the Parabola
We start with the vertex form of a quadratic equation:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\( (h, k) \)[/tex] is the vertex. Here, [tex]\( h = -4 \)[/tex] and [tex]\( k = 2 \)[/tex]. Substituting these values, we get:
[tex]\[ y = a(x + 4)^2 + 2 \][/tex]
### Step 2: Use the y-intercept to Find [tex]\( a \)[/tex]
We are given the [tex]\( y \)[/tex]-intercept [tex]\((0, -30)\)[/tex]. This point is on the parabola, so we can substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = -30 \)[/tex] into the equation to solve for [tex]\( a \)[/tex]:
[tex]\[ -30 = a(0 + 4)^2 + 2 \][/tex]
[tex]\[ -30 = a(16) + 2 \][/tex]
[tex]\[ -30 - 2 = 16a \][/tex]
[tex]\[ -32 = 16a \][/tex]
[tex]\[ a = -2 \][/tex]
### Step 3: Write the Complete Equation
Now that we know [tex]\( a = -2 \)[/tex], we can write the full equation of the parabola:
[tex]\[ y = -2(x + 4)^2 + 2 \][/tex]
### Step 4: Find the [tex]\( x \)[/tex]-intercepts
The [tex]\( x \)[/tex]-intercepts occur where [tex]\( y = 0 \)[/tex]. Setting [tex]\( y = 0 \)[/tex] in the equation and solving for [tex]\( x \)[/tex] we get:
[tex]\[ 0 = -2(x + 4)^2 + 2 \][/tex]
[tex]\[ -2(x + 4)^2 + 2 = 0 \][/tex]
[tex]\[ -2(x + 4)^2 = -2 \][/tex]
[tex]\[ (x + 4)^2 = 1 \][/tex]
Taking the square root of both sides:
[tex]\[ x + 4 = 1 \quad \text{or} \quad x + 4 = -1 \][/tex]
[tex]\[ x = 1 - 4 = -3 \quad \text{or} \quad x = -1 - 4 = -5 \][/tex]
### Step 5: Verify the Intercepts
The solutions to the equation are [tex]\( x = -3 \)[/tex] and [tex]\( x = -5 \)[/tex].
Thus, the [tex]\( x \)[/tex]-intercepts of the parabola are:
[tex]\[ (-5, 0) \quad \text{and} \quad (-3, 0) \][/tex]
### Final Answer
[tex]\[ \left(x_1, y_1\right), \left(x_2, y_2\right) = (-5, 0), (-3, 0) \][/tex]
