A parabola with vertex (2,0) and axis of symmetry parallel to the y-axis, passes through (3,1) and (-3,t). Find the value of t.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let the parabola be :

f(x)= ax^2+ bx + c

The vertex is ( 2,0)

Then:

Vx = -b/2a = 2

==> -b =4a

==> b= -4a ............(1)

Vy = (b^2-4ac) / 4a = 0

==> b^2 - 4ac = 0

==> (-4a)^2 - 4ac = 0

==> 16a^2 = 4ac

==> 4a = c

==> -4a = -c

==> b = -c ............(2)

==> f(x) = ax^2 -4a x + 4a

Now we will substitute with (3,1)

==> 1= 9a -4a*3 + 4a

==> 1= 9a - 12a + 4a

=> 1= a

==> a = 1

==> b= -4a= -4

==> c= 4a= 4

==> f(x) = x^2 -4x + 4

==>Now let us substitute ( -3,t)

==> t= (-3)^2 -4(-3) + 4

==> t= 9 + 12 + 4 = 25

=> t= 25

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We first need to find the equation of the parabola that has a vertex (2,0). Now the equation of a parabola with vertex ( h, k) is given as

y= a(x-h)^2 + k.

=> y = a(x- 2)^2 + 0.

Now the parabola passes through (3,1)

=> 1 = a( 3 - 2)^2

=> 1 = a

Therefore the equation of the parabola is y = (x - 2)^2

Now as the point (-3 , t) lies on the parabola

t = (-3 - 2)^2

=> t = 5^2

=> t = 25

Therefore the required value of t is 25.

changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

Hi.

I have verified the solution by the others by plotting the answer at graphsketch.com

Click on the link at the bottom of the answer to appreciate the question graphically.

Sources:

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