If Vertex is (4,-7) and x intercept is 4, what is the standard equation of the parabola?
no calculus need for this. Remember your Algebra I
Vertex is (4,-7) ... y = a(x-4)^2 - 7
But the x-intercept cannot be 4, since we know that y(4) = -7
However, if you meant that the y-intercept is 4, then
a(0-4)^2 - 7 = 7
a = 7/8
and thus
y = 7/8 (x-4)^2 - 7
Now just rearrange that into standard form.
To find the standard equation of a parabola, we need to know the vertex form of the equation. The vertex form of a parabola is given by:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola.
Given:
Vertex = (4, -7)
x-intercept = 4
Since the x-intercept is the point where the parabola crosses the x-axis, this means that when y = 0, x = 4.
Let's substitute these values into the vertex form equation:
0 = a(4 - 4)^2 + (-7)
Simplifying the equation:
0 = a(0)^2 + (-7)
0 = 0 + (-7)
0 = -7
Since the equation simplifies to 0 = -7, this means that there is no value of 'a' that will satisfy this equation.
Therefore, there is no standard equation of the parabola that satisfies the given conditions.
To find the standard equation of a parabola, we need to use the vertex form of a parabola which is given by:
y = a(x-h)^2 + k
Where (h, k) represents the vertex of the parabola.
Given that the vertex is (4, -7), we have h = 4 and k = -7. Plugging these values into the equation, we get:
y = a(x-4)^2 - 7
Now, we need to find the value of 'a' to complete the equation.
We are also given that the x-intercept of the parabola is 4. The x-intercept occurs when y = 0. Plugging this into the equation, we have:
0 = a(4-4)^2 - 7
0 = a(0)^2 - 7
0 = -7
Since a parabola cannot have a slope of 0, this means that there is no value of 'a' that satisfies the equation. Therefore, the standard equation of the parabola cannot be determined.
In summary, the given information does not allow us to find the standard equation of the parabola, as it cannot be determined without the value of 'a'.