Answer:
[tex] \boxed{ \boxed{ \mathfrak{a = - \frac{1}{2 } \: b = 2 \: c = 4}}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
tips and formulas:
- the parabola crosses f(x) at (4,0)
- the vertex of the parabola (2,6)
- vertex:(h,k)
- h=-b/2a
- k=f(x)
- vertex form:f(x)=a(x-h)²+k
- standard form:ax²+bx+c
let's solve:
the vertex form of the equation is
- [tex] \sf f(x) = a(x - 2 {)}^{2} + 6[/tex]
let's figure out a
since the parabola crosses f(x) at (0,4)
therefore
- [tex] \sf a(0 - {2)}^{2} + 6 = 4[/tex]
- [tex] \sf simplify \: parentheses : \\ \sf a( - 2 {)}^{2} + 6 = 0[/tex]
- [tex] \sf simplify \: squres : \\ \sf \: 4a + 6 = 4[/tex]
- [tex] \sf cancel \: 6 \: from \: both \: sides : \\ \sf 4a + 6 - 6 = 4 - 6 \\ \sf 4a = - 2 [/tex]
- [tex] \sf divide \: both \: sides \: by \: 4 : \\ \sf \frac{4a}{4} = - \frac{ 2}{4} \\ \sf a = - \frac{1}{2} [/tex]
let's figure out b,c
- we have to substitute the value of a into the vertex form and then simply it to get b and c
- [tex] \sf substitue \: the \: value \: of \: a \: into \: the \: vertex \: form \: of \: the \: equation : \\ \sf \frac{1}{2} (x - 2 {)}^{2} + 6[/tex]
- [tex] \sf use \: (a - b {)}^{2} = {a}^{2} - 2ab + {b}^{2} \: to \: simplify \: the \: exponet : \\ \sf - \frac{1}{2} ( {x}^{2} - 2.x.2 + {2}^{2} ) + 6 \\ \sf - \frac{1}{2} ( {x}^{2} - 4x + 4) + 6[/tex]
- [tex] \sf distribut e \: - \frac{1}{2} \: between \: the \: parentheses : \\ \sf - \frac{1}{2} x + 2x - 2 + 6[/tex]
- [tex] \sf add : \\ \sf - \frac{1}{2} x + 2x + 4[/tex]
therefore
- [tex] \sf \: b = 2[/tex]
- [tex] \sf \: c = 4[/tex
