Answer:
[tex]a=8[/tex]
[tex]b=0[/tex]
[tex]c=1[/tex]
Step-by-step explanation:
The vertex form of a quadratic is [tex]y=a(x-h)^2+k[/tex] where the vertex is [tex](h,k)[/tex].
We are given [tex]h=0,k=1[/tex].
Plugging this in gives us:
[tex]y=a(x-0)^2+1[/tex]
Simplifying:
[tex]y=ax^2+1[/tex]
Since it passes through [tex](x,y)=(1,9)[/tex], we can use this information to find [tex]a[/tex].
[tex]9=a(1)^2+1[/tex]
[tex]9=a+1[/tex]
[tex]8=a[/tex]
So the quadratic in vertex form is [tex]y=8x^2+1[/tex].
That happens to be in standard form as well.
[tex]a=8[/tex]
[tex]b=0[/tex]
[tex]c=1[/tex]
The equation of a parabola is [tex]y = a(x - h)^2 + k[/tex], where (h,k) represents the vertex.
The values of a, b and c are 8, 0 and 1, respectively.
We have:
[tex](h,k) = (0,1)[/tex] --- the vertex
[tex](x,y) = (1,9)[/tex] --- the point it passes through
Substitute [tex](h,k) = (0,1)[/tex] in [tex]y = a(x - h)^2 + k[/tex]
[tex]y = a(x - 0)^2 + 1[/tex]
[tex]y = ax^2 + 1[/tex]
Substitute [tex](x,y) = (1,9)[/tex] in [tex]y = ax^2 + 1[/tex]
[tex]9= a \times 1^2 +1[/tex]
[tex]9= a \times 1 +1[/tex]
[tex]9= a +1[/tex]
Collect like terms
[tex]a= 9 -1[/tex]
[tex]a = 8[/tex]
Substitute [tex]a = 8[/tex] in [tex]y = ax^2 + 1[/tex]
So, the equation of the parabola is:
[tex]y = 8x^2 + 1[/tex]
Compare [tex]y = 8x^2 + 1[/tex] to [tex]f(x)=ax^2+bx+c[/tex]
[tex]a = 8\\b = 0\\c =1[/tex]
See attachment for the equation of [tex]y = 8x^2 + 1[/tex]
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