Find the quadratic function that has vertex (-1, 8) and whose graph goes through the point (8, 170).
I could not understand this problem I Can not for the lift of me get it. I got as far as some number(x+1)^2+8 but i don't know what the number before all that would be for the vertical stretch
remember the form, y = a(x-h)^2 + k
where (h,k) is the vertex, and x and y are points on the graph,,
therefore, solving for a,
170 = a*(8-(-1))^2 + 8
170 = a*81 + 8
81a = 162
a = 2
therefore,
y = 2(x+1)^2 + 8
expanding this,
y = 2x^2 + 4x + 10
so there,, :)
Thank you so much!!!
To find the quadratic function that has a given vertex and passes through a given point, we need to use the standard form of a quadratic function: f(x) = a(x - h)^2 + k, where (h, k) represents the vertex.
In this case, the vertex is (-1, 8). So, h = -1 and k = 8.
Plugging these values into the standard form equation, we have: f(x) = a(x - (-1))^2 + 8
Simplifying, f(x) = a(x + 1)^2 + 8.
Now, let's use the given point (8, 170) to find the value of 'a':
When the point (8, 170) is plugged into the equation, we have:
170 = a(8 + 1)^2 + 8
Simplifying further, we have:
170 = a(9)^2 + 8
170 = 81a + 8
162 = 81a
a = 2
Now we have the value of 'a' (a = 2), so we can substitute it back into the equation:
f(x) = 2(x + 1)^2 + 8
Therefore, the quadratic function that has a vertex (-1, 8) and passes through the point (8, 170) is f(x) = 2(x + 1)^2 + 8.