vertex form of a quadratic equation

Question

A certain quadratic equation is graphed on an xy-coordinate plane. The graph of the equation forms a parabola that opens upwards and has its vertex at the point (3, -4). One of the x-intercepts of the graph is at x = 2. What is the equation of the parabola in vertex form?

Answer 1

The vertex form of a parabola takes the form of y = a*(x-h)² + k, where h is the x-coordinate of the vertex, k is the y-coordinate of the vertex, and a is a constant.

In this problem, we are given the vertex coordinates. The x-coordinate of the vertex is 3, and the y-coordinate of the vertex is -4. Therefore, h = 3 and k = -4.

Replacing these values into the equation gives y = a*(x-3)² - 4. We need to find a still. We are given the information that the x-intercept ocurrs at x = 2. Also, from the very definition of an x-intercept, we know that y=0 for x=2. This means the function contains the point (2,0). Now, we can replace x = 2 and y = 0 into the equation to find a.

0 = a * (2-3)² - 4

0 = a * (-1)² - 4

0 = a - 4

a = 4

The final answer is y = 4 * (x-3)² - 4

Answer 2

In vertex form, a quadratic equation is represented as y = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.

We know from the problem that h = 3 and k = -4. So, the equation becomes y = a(x-3)^2 - 4.

Given that one of the x-intercepts is at x = 2, we can substitute this value into the equation (since the y-value is 0 at x-intercepts) and solve for 'a'.

0 = a(2-3)^2 - 4,

0 = a(1)^2 - 4,

0 = a - 4,

a = 4.

So, the equation of the parabola is y = 4(x-3)^2 - 4.