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The graph of
y = x^2+px+q has a turning point at (2,5) Find the values of p and q

1 Answer

4 votes

Answer:

p = -4

q = 9

Explanation:

The vertex form of a parabola is:

y = a (x − h)² + k

Given that the leading coefficient is 1 and the vertex is (2,5):

y = (x − 2)² + 5

Distributing:

y = x² − 4x + 4 + 5

y = x² − 4x + 9

Alternatively, using calculus, if y = x² + px + q, then the derivative is:

dy/dx = 2x + p

The turning point occurs when the derivative is 0:

0 = 2x + p

0 = 2(2) + p

p = -4

Plugging in and solving for q:

y = x² − 4x + q

5 = (2)² − 4(2) + q

5 = 4 − 8 + q

q = 9

User Kit Ramos
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