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Retired Moderator
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WE:Education (Education)
The graph of a quadratic function f(x) passes through (9, 0), (0, 9),
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14 Aug 2021, 12:07
14 Aug 2021, 12:07
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The graph of a quadratic function f(x) passes through (9, 0), (0, 9), and (7, 30). What is the equation of axis of symmetry of f(x)?
Show: ::
\(x = \frac{17}{4}\)
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
The graph of a quadratic function f(x) passes through (9, 0), (0, 9),
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15 Aug 2021, 06:26
15 Aug 2021, 06:26
1
\(f(x) = ax^2 + bx + c\)
Plugging-in (0, 9);
\(9 = a(0)^2 + b(0) + c\)
\(c = 9\)
So, \(f(x) = ax^2 + bx + 9\)
Plugging-in (9, 0);
\(0 = a(9)^2 + b(9) + 9\)
\(81a + 9b = -9\)
\(9a + b = -1\) ....... (1)
Plugging-in (7, 30);
\(30 = a(7)^2 + b(7) + 9\)
\(49a + 7b = 21\)
\(7a + b = 3\) ....... (2)
Subtracting (1) and (2);
\(9a + b = -1\)
\(7a + b = 3\)
\(2a = -4\)
\(a = -2\)
Substituting \(a = -2\) in (2);
\(7(-2) + b = 3\)
\(b = 17\)
Finally, \(f(x) = -2x^2 + 17x + 9\)
Axis of symmetry is given by \(x = x_v\), where \(x_v\) is the coordinate of the vertex
\(x_v = \frac{-b}{2a} = \frac{-17}{2(-2)} = \frac{17}{4}\)
Plugging-in (0, 9);
\(9 = a(0)^2 + b(0) + c\)
\(c = 9\)
So, \(f(x) = ax^2 + bx + 9\)
Plugging-in (9, 0);
\(0 = a(9)^2 + b(9) + 9\)
\(81a + 9b = -9\)
\(9a + b = -1\) ....... (1)
Plugging-in (7, 30);
\(30 = a(7)^2 + b(7) + 9\)
\(49a + 7b = 21\)
\(7a + b = 3\) ....... (2)
Subtracting (1) and (2);
\(9a + b = -1\)
\(7a + b = 3\)
\(2a = -4\)
\(a = -2\)
Substituting \(a = -2\) in (2);
\(7(-2) + b = 3\)
\(b = 17\)
Finally, \(f(x) = -2x^2 + 17x + 9\)
Axis of symmetry is given by \(x = x_v\), where \(x_v\) is the coordinate of the vertex
\(x_v = \frac{-b}{2a} = \frac{-17}{2(-2)} = \frac{17}{4}\)
General Discussion
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
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GRE 1: Q160 V152
The graph of a quadratic function f(x) passes through (9, 0), (0, 9),
[#permalink]
14 Aug 2021, 13:33
14 Aug 2021, 13:33
1
so, it's y-intercept is 9 from (0,9) and one of the roots is 9 from (9,0)
\((x-9)(x-1)=0\) is the present equation's root form and it's quadratic equation form is \(x^2-10x+9=0\)
let's check equation with pair (7, 30) ---> 49-70+9=-12
did not work
also, the axis of symmetry is a vertical line x=−b/2a, and spoiler suggests x=17/4
then quadratic equation must be in the form \(2x^2-17x+9=0\)
plugging (7,30) into the above equation derived from the axis of symmetry equation, results in \(2*49-17*7+9=-12\) not 30
\((x-9)(x-1)=0\) is the present equation's root form and it's quadratic equation form is \(x^2-10x+9=0\)
let's check equation with pair (7, 30) ---> 49-70+9=-12
did not workalso, the axis of symmetry is a vertical line x=−b/2a, and spoiler suggests x=17/4
then quadratic equation must be in the form \(2x^2-17x+9=0\)
plugging (7,30) into the above equation derived from the axis of symmetry equation, results in \(2*49-17*7+9=-12\) not 30
KarunMendiratta wrote:
The graph of a quadratic function f(x) passes through (9, 0), (0, 9), and (7, 30). What is the equation of axis of symmetry of f(x)?
Show: ::
\(x = \frac{17}{4}\)
