A quadratic function has its vertex at the point (-5,-2). The function passes through the point (-8,-10). Find the quadratic and linear coefficients and the constant term of the function.

clearly, y = a(x+5)^2 - 2

Now just plug in the point and solve for a. But, you want y=ax^2+bx+c, so just expand and collect terms to wind up in that format.

To find the quadratic and linear coefficients and the constant term of the quadratic function, we can use the vertex form of a quadratic function:

y = a(x - h)^2 + k,

where (h, k) represents the vertex of the parabola. In this case, the vertex is (-5, -2), so we can substitute these values into the vertex form to get:

y = a(x - (-5))^2 + (-2),
y = a(x + 5)^2 - 2.

Now, we need to use the given point (-8, -10) to determine the value of 'a'. Substituting these coordinates into the equation, we have:

-10 = a(-8 + 5)^2 - 2,
-10 = a(-3)^2 - 2,
-10 = 9a - 2,
9a = -10 + 2,
9a = -8,
a = -8/9.

Therefore, the quadratic coefficient is a = -8/9.
The linear coefficient is 0 since there is no x term.
The constant term is found by substituting the value of 'a' into the vertex form equation:

y = (-8/9)(x + 5)^2 - 2.
So, the constant term is -2.

In summary, the quadratic coefficient is -8/9, the linear coefficient is 0, and the constant term is -2.

To find the quadratic function with a vertex at (-5,-2) and passing through (-8,-10), we can use the standard form of a quadratic function:

f(x) = ax^2 + bx + c

Since the vertex is given by (-5, -2), we know that the x-coordinate of the vertex is -5. This means that the equation of the parabola can be written in the form:

f(x) = a(x + 5)^2 + k

We need to find the value of 'a' and 'k'.

Plugging in the coordinates of the point (-8,-10) into the equation, we get:

-10 = a(-8 + 5)^2 + k

Next, we can substitute the y-coordinate of the vertex (-2) for 'k':

-10 = a(-8 + 5)^2 - 2

Now, let's simplify the equation:

-10 = a(-3)^2 - 2
-10 = a(9) - 2
-10 = 9a - 2

Now, let's solve for 'a':

9a - 2 = -10
9a = -8
a = -8/9

So, the quadratic coefficient is -8/9.

Now, let's find the value of 'k' using the vertex point (-5, -2):

-2 = -8/9(-5 + 5)^2 + k
-2 = k

So, the constant term is -2.

Finally, let's find the value of 'b' by expanding the equation:

f(x) = a(x + 5)^2 + k
f(x) = -8/9(x + 5)^2 - 2
f(x) = -8/9(x^2 + 10x + 25) - 2
f(x) = -8/9x^2 - 80/9x - 200/9

Therefore, the quadratic and linear coefficients of the quadratic function are -8/9 and -80/9 respectively, and the constant term is -200/9.