Can you please help me with the below question:
The vertex of a parabola is located at (-12, -1). The parabola also passes through the point (-10, 5).
Write the equation of this parabola in both vertex form and standard form.
Vertex form:
Standard form:
Thank you
using the vertex:
y = a(x+12)^2 - 1
but (-10,5) lies on it, so ...
5= a(2^2) - 1
6 = 4a
a = 3/2
y = (3/2)(x + 12)^2 - 1
expand to get the other form
To find the equation of the parabola in both vertex form and standard form, we need to use the given information about the vertex and another point that lies on the parabola.
First, let's start with the vertex form of the equation. The general equation of a parabola in vertex form is given as:
y = a(x - h)^2 + k
where (h, k) represents the coordinates of the vertex.
In this case, the vertex is given as (-12, -1), so we have:
y = a(x - (-12))^2 + (-1)
y = a(x + 12)^2 - 1
Now, to determine the value of 'a', we can use the other point (-10, 5) that lies on the parabola. Since this point satisfies the equation, we can substitute the values of x and y to solve for 'a':
5 = a((-10) + 12)^2 - 1
Simplifying this equation, we get:
5 = 4a - 1
4a = 6
a = 6/4
a = 3/2
Now that we know the value of 'a', we can substitute it back into the vertex form equation to get the final equation:
y = (3/2)(x + 12)^2 - 1
This is the equation of the parabola in vertex form.
Next, let's convert the equation into standard form, which is in the form of:
y = ax^2 + bx + c
To convert the equation, we expand the squared term:
y = (3/2)(x^2 + 24x + 144) - 1
y = (3/2)x^2 + 36x + 216 - 1
y = (3/2)x^2 + 36x + 215
This is the equation of the parabola in standard form.
To summarize:
Vertex form: y = (3/2)(x + 12)^2 - 1
Standard form: y = (3/2)x^2 + 36x + 215