Find f(5), if the graph of the quadratic function f(x)=ax^2+bx+c intersects the ordinate axis at point (0;3) and its vertex is at point (2;0)

Paloma Sanford

Paloma Sanford

Answered question

2022-10-20

Find f(5), if the graph of the quadratic function f ( x ) = a x 2 + b x + c intersects the ordinate axis at point (0;3) and its vertex is at point (2;0)
So I used the vertex form, y = ( x 2 ) 2 + 3, got the quadratic equation and then put 5 instead of x to get the answer, but it's wrong. I think I shouldn't have added 3 in the vertex form but I don't know how else I can solve this

Answer & Explanation

Reese Hobbs

Reese Hobbs

Beginner2022-10-21Added 13 answers

Step 1
You are right that one solution is to use the vertex form
y = a ( x h ) 2 + k
but you have k wrong. If the vertex is at (2,0) then h = 2 (which you have) and k = 0 (which you got wrong). So the equation is now
y = a ( x 2 ) 2 + 0
or
y = a ( x 2 ) 2
Now use the fact that (0,3) is on the parabola, so x = 0 , y = 3 satisfies the equation:
3 = a ( 0 2 ) 2
So a = 3 4 . Thus the full equation is
y = 3 4 ( x 2 ) 2
Step 2
Finally, we want the value of y when x = 5, which is
y = 3 4 ( 5 2 ) 2
or
y = 27 4
mafalexpicsak

mafalexpicsak

Beginner2022-10-22Added 4 answers

Step 1
As we know: 2 a x + b = 0 when x = 2
This means 4 a + b = 0 or b = 4 a
Step 2
Now you can substitute
a × 2 2 4 a × 2 + 3 = 0
Or 4 a 8 a + 3 = 0
Or a = 3 4
Now you can get b and you can get f(5)

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