\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
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Convert the function \(f(x) = x^2 + 6x + 5\) to vertex form, \(f(x)=a(x-h)^2+k\).
\(f(x)=\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
Then, identify the coefficients.
\(a=\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
, \(h=\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
and \(k=\)
\(\pi\)
\(e\)
\(x\)
\(n\)
\(u_n\)
\(f\)
\(i\)
\(\frac{a}{b}\)
\(\sqrt{\,}\)
\({a}^{b}\)
\(\ln{\,}\)
\(\log{\,}\)
!
\(C\)
7
8
9
←
→
\(\sin{\,}\)
4
5
6
(
)
\(\cos{\,}\)
1
2
3
\(\times\)
\(\div\)
\(\tan{\,}\)
C
0
.
+
-
=
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