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Difficulty: MediumQuadratic Functions and Graphs

The graph of the quadratic function f(x)=x2−4x+7f(x) = x^2 - 4x + 7 is translated 33 units to the right and 22 units down in the xyxy-plane to form the graph of the function g(x)=x2+px+qg(x) = x^2 + px + q, where pp and qq are constants. What is the value of qq?

Answer: 26

Answer

The value of qq is 2626.
To find the constant term qq of the translated quadratic function g(x)=x2+px+qg(x) = x^2 + px + q, we first determine the vertex of the original function f(x)=x2−4x+7f(x) = x^2 - 4x + 7. Completing the square gives f(x)=(x−2)2+3f(x) = (x - 2)^2 + 3, which identifies the vertex of the parabola as (2,3)(2, 3). Translating the graph 33 units to the right and 22 units down shifts the vertex to (2+3,3−2)=(5,1)(2 + 3, 3 - 2) = (5, 1). Because the translation does not affect the shape of the parabola, the leading coefficient remains 11. The vertex form of the new function is g(x)=(x−5)2+1g(x) = (x - 5)^2 + 1. Expanding this expression yields g(x)=x2−10x+25+1=x2−10x+26g(x) = x^2 - 10x + 25 + 1 = x^2 - 10x + 26. Comparing this to g(x)=x2+px+qg(x) = x^2 + px + q, we find that q=26q = 26.

Step-by-Step Solution

1
Convert the original function f(x)=x2−4x+7f(x) = x^2 - 4x + 7 into vertex form, f(x)=a(x−h)2+kf(x) = a(x - h)^2 + k, to identify its vertex (h,k)(h, k).
f(x)=(x−2)2+3f(x) = (x - 2)^2 + 3, which represents a parabola with vertex (2,3)(2, 3).
Finding the vertex of the original function allows us to apply the translation directly to the vertex coordinates.
2
Apply the translation of 33 units to the right and 22 units down to the coordinates of the vertex (2,3)(2, 3).
The new vertex is (2+3,3−2)=(5,1)(2 + 3, 3 - 2) = (5, 1).
Translating a graph shifts its vertex by the corresponding horizontal and vertical amounts.
3
Write the equation of the translated function g(x)g(x) in vertex form using the new vertex (5,1)(5, 1) and the original leading coefficient a=1a = 1.
g(x)=(x−5)2+1g(x) = (x - 5)^2 + 1
A translation does not change the shape or vertical stretch of the parabola, so the coefficient of x2x^2 remains 11.
4
Expand the vertex form of g(x)g(x) into standard form, g(x)=x2+px+qg(x) = x^2 + px + q, to determine the constant term qq.
g(x)=x2−10x+26g(x) = x^2 - 10x + 26, which means q=26q = 26.
Expanding the equation allows us to compare it directly with the standard form of g(x)g(x) and identify the value of the constant term.

Key Concept

Vertex form and translations of quadratic functions

Alternative Method

Alternatively, the translation can be applied directly to the variable xx in the function equation. Translating a function f(x)f(x) by 33 units to the right and 22 units down yields g(x)=f(x−3)−2g(x) = f(x - 3) - 2. Substituting x−3x - 3 into the original function gives: g(x)=(x−3)2−4(x−3)+7−2g(x) = (x - 3)^2 - 4(x - 3) + 7 - 2. Simplifying this expression: g(x)=(x2−6x+9)−(4x−12)+5=x2−10x+26g(x) = (x^2 - 6x + 9) - (4x - 12) + 5 = x^2 - 10x + 26. This directly shows that the constant term qq is 2626.
Estimated Time:1m 30s
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Topics

Quadratic Functions and Graphs

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