A quadratic function has a vertex of (3,-6) and the point (-1,10) lies on the graph of the function. Write the function in standard form.
If you could help me with this question, that would be great. Thanks in advance.
using the vertex form,
y = a(x-3)^2 - 6
using the point given,
10 = a(-1-3)^2 - 6
10 = 16a-6
a = 1
y = (x-3)^2 - 6 = x^2-6x+3
Which quadratic function in vertex form can be represented by the graph that has a vertex at (3,-7) and passes through the point (1,-10)
Well, I'll certainly do my best to help you with this question! It sounds pretty quadratic-tastic.
First, let's use the vertex form of a quadratic function, which is f(x) = a(x - h)^2 + k, where (h, k) is the vertex.
We know that the vertex is (3, -6), so we can plug that into the equation to get f(x) = a(x - 3)^2 - 6.
Now, let's use the point (-1, 10) to find the value of a. Plug in this point into the equation and solve for a:
10 = a(-1 - 3)^2 - 6
10 = a(-4)^2 - 6
10 = 16a - 6
16a = 16
a = 1
So now we have the equation f(x) = 1(x - 3)^2 - 6.
To get the function in standard form, we'll expand the squared term:
f(x) = 1(x^2 - 6x + 9) - 6
And then simplify:
f(x) = x^2 - 6x + 9 - 6
f(x) = x^2 - 6x + 3
And there you have it! The function in standard form is f(x) = x^2 - 6x + 3.
I hope that makes you smile, even if quadratic functions can be a bit grumpy sometimes!
To write a quadratic function in standard form, we need the vertex form of the function, which is given by:
f(x) = a(x-h)^2 + k
where (h, k) represents the coordinates of the vertex.
From the given information, we know that the vertex is (3,-6). Therefore, our function can be written as:
f(x) = a(x-3)^2 - 6
The next step is to find the value of 'a'. To do this, we can use the fact that the point (-1,10) lies on the graph of the function.
Substituting the coordinates of the point into the function, we get:
10 = a(-1-3)^2 - 6
Simplifying this equation, we have:
10 = a(-4)^2 - 6
10 = a(16) - 6
10 = 16a - 6
16a = 10 + 6
16a = 16
a = 16/16
a = 1
Now that we have the value of 'a', we can substitute it back into the function:
f(x) = (x-3)^2 - 6
Therefore, the quadratic function in standard form with a vertex of (3,-6) and the point (-1,10) lying on the graph is:
f(x) = (x-3)^2 - 6.
To find the quadratic function in standard form given the vertex and a point, we can use the vertex form of a quadratic function and then convert it to standard form.
The vertex form of a quadratic function is given by:
f(x) = a(x - h)^2 + k
Where (h, k) is the vertex of the parabola.
We are given that the vertex of the quadratic function is (3, -6). So h = 3 and k = -6.
Substituting these values into the equation, we get:
f(x) = a(x - 3)^2 - 6
Now, we need to use the given point (-1, 10) to solve for the value of "a".
Substituting the coordinates of the point (-1, 10) into the equation, we get:
10 = a(-1 - 3)^2 - 6
Simplifying this equation:
10 = a(-4)^2 - 6
10 = 16a - 6
16a = 10 + 6
16a = 16
a = 1
Now that we have the value of "a", we can substitute it back into the equation:
f(x) = 1(x - 3)^2 - 6
Expanding and simplifying this equation:
f(x) = (x - 3)^2 - 6
To convert this equation into standard form, we need to expand the squared term:
f(x) = (x - 3)(x - 3) - 6
f(x) = x^2 - 6x + 9 - 6
f(x) = x^2 - 6x + 3
So, the function in standard form is:
f(x) = x^2 - 6x + 3
Therefore, the quadratic function in standard form that satisfies the given conditions is f(x) = x^2 - 6x + 3.