Which quadratic function has its vertex at (2,3) and passes through (1,0)?
a. y = 2(x-2)^2+3
b. y = -3(x-2)^2+3
c. y = -3(x+2)^2+3
d. y = 2(x-2)^2-3
Could you explain why you got your answer, too?
To determine the quadratic function that satisfies the given conditions, we need to use the standard form of a quadratic equation: y = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.
Given that the vertex is (2, 3), this information tells us that h = 2 and k = 3. Plugging these values into the equation gives us y = a(x-2)^2 + 3.
Next, we need to determine the value of 'a'. To find 'a', we can use the point (1, 0) which lies on the parabola.
Substituting x = 1 and y = 0 into the equation y = a(x-2)^2 + 3, we get:
0 = a(1-2)^2 + 3
0 = a(-1)^2 + 3
0 = a + 3
Solving the equation, we find that a = -3.
Therefore, the quadratic function that satisfies the given conditions is y = -3(x-2)^2 + 3, which corresponds to option b.
To find the quadratic function that has its vertex at (2,3) and passes through (1,0), we can use the vertex form of a quadratic function, which is given by:
y = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex.
In this case, the vertex is given as (2,3), so h = 2 and k = 3. Substituting these values into the vertex form equation, we have:
y = a(x - 2)^2 + 3
Now we need to determine the value of 'a'. Since the function passes through the point (1,0), we can substitute these coordinates into the equation:
0 = a(1 - 2)^2 + 3
Simplifying the equation, we have:
0 = a(-1)^2 + 3
0 = a + 3
Subtracting 3 from both sides, we get:
-3 = a
Now, substitute this value of 'a' back into the equation, we have:
y = -3(x - 2)^2 + 3
Therefore, the correct answer is option (c): y = -3(x+2)^2+3.
Explanation:
- We started by using the vertex form of a quadratic function, which helped us represent the equation with the known vertex coordinates.
- By substituting the given vertex coordinates (2,3) into the vertex form equation, we obtained an equation with unknown 'a' value.
- Next, using the fact that the function passes through the point (1,0), we substituted these coordinates into the equation to find the value of 'a'.
- Finally, by substituting the found 'a' value back into the equation, we obtained the correct quadratic function (c).
if y = a(x-p)^2 + q then the vertex is (p,q)
both a) and b) match that pattern
but which one does (1,0) satisfy?
substitute to find out.