In my homework I am given a parabola and it says to give the equation of the parabola in standard form. It gives me the Vertex, the y intercept and the zeros. I understand that y=ax2 +bx+c. C is...
In my homework I am given a parabola and it says to give the equation of the parabola in standard form. It gives me the Vertex, the y intercept and the zeros.
I understand that y=ax2 +bx+c. C is the y intercept, but I do not know how to get a or b. Please help me understand.
Let us consider a specific example: Given the vertex of the parabola is at (3,-2), the y-intercept is (0,16), and the zeros are 2 and 4 find the equation of the quadratic in standard form.
Note that a quadratic function in one variable can be written in many forms:
(1) Intercept or factored form y=a(x-p)(x-q) where p,q are the zeros of the function.
(2) Vertex form `y=a(x-h)^2+k ` where the vertex is at (h,k).
(3) Standard form `y=ax^2+bx+c ` : here c is the y-intercept, and the x-coordinate of the vertex is `x=(-b)/(2a) ` .
We can move from one form to another using algebraic techniques -- to get from factored form to standard form you multiply the factors. To get from standard form to vertex form you can complete the square. To get from vertex form to standard form you expand the binomial term, etc...
If you are given the vertex, the zeros, and the y-intercept there are many ways to get to the standard form.
(1) Using the example above: find the factored form and multiply to get the standard form. Since 2 and 4 are zeros we have y=a(x-2)(x-4). Since (0,16) is a point on the curve, substitute for x and y to solve for a:
16=a(0-2)(0-4) ==> a=2 so y=2(x-2)(x-4). Now multiply to get:
Note that we could have used the point (3,-2) to solve for a, but it is computationally easier to use the intercept since x=0.
(2) Again using the example above, find the vertex form and then expand to get the standard form.
The vertex is (3,-2) so h=3 and k=-2. So `y=a(x-3)^2-2 ` . Use a known point to solve for a: we can use any of (0,16),(2,0), or (4,0). Using (4,0) we substitute 4 for x and 0 for y to get:
`0=a(4-3)^2-2 ==> a=2 `
Then `y=2(x-3)^2-2=2[x^2-6x+9]-2=2x^2-12x+16 `
(3)We only need three noncollinear points to determine a parabola (assuming it is a function.). Here you know (3,-2),(2,0) and (4,0). So you could develop and solve simultaneously three linear equations of the form `y=ax^2+bx+c ` for the coefficients a,b, and c.
` a(0)^2+b(0)+c=16 ==> c=16 `
`a(2)^2+b(2)+16=0 ==> 4a+2b=-16 ==> 2a+b=-8 `
`a(4)^2+b(4)+16=0 ==> 16a+4b=-16 ==> 4a+b=-4 `
Then solving the last two equations we get 2a=4 or a=2 and b=-12.
So `y=2x^2-12x+16 ` as before.
There are other methods.