Question

2.4.4 quiz: parabolas with vertices not at the origin. the vertex of this parabola is at (-3, -1). when the y - value is 0, the x - value is 4. what is the coefficient of the squared term in the parabolas equation? a. -3 b. 3 c. -7 d. 7

Explanation:

Step1: Write the vertex - form of parabola equation

The vertex - form of a parabola with vertex \((h,k)\) is \(y=a(x - h)^2+k\). Here, \(h=-3\) and \(k = - 1\), so the equation is \(y=a(x + 3)^2-1\).

Step2: Substitute the given point into the equation

We know that when \(y = 0\), \(x = 4\). Substitute these values into the equation \(0=a(4 + 3)^2-1\).

Step3: Solve for \(a\)

First, simplify the equation: \(0=a\times49-1\). Then, add 1 to both sides: \(1 = 49a\). Divide both sides by 49, we get \(a=\frac{1}{49}\). But if we assume the parabola is of the form \(x=a(y - k)^2+h\) (since the parabola seems to open horizontally from the graph - like appearance), with \(h=-3,k = - 1\), the equation is \(x=a(y + 1)^2-3\). Substitute \(x = 4,y = 0\) into it: \(4=a(0 + 1)^2-3\). Add 3 to both sides: \(7=a\).

Answer:

D. 7