Sure, let's determine which of the given equations represents a parabola with a vertex at the point [tex]\((-3, 2)\)[/tex].
For a quadratic equation of the form [tex]\(y = ax^2 + bx + c\)[/tex], the vertex [tex]\((h, k)\)[/tex] can be found using the formulas:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = c - \frac{b^2}{4a} \][/tex]
We need to find which equation satisfies [tex]\( h = -3 \)[/tex] and [tex]\( k = 2 \)[/tex].
1. Consider the equation [tex]\(y = 4x^2 + 24x + 38\)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 24 \)[/tex]
- [tex]\( c = 38 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{24}{2 \times 4} = -\frac{24}{8} = -3 \][/tex]
Calculate [tex]\( k \)[/tex]:
[tex]\[ k = c - \frac{b^2}{4a} = 38 - \frac{24^2}{4 \times 4} = 38 - \frac{576}{16} = 38 - 36 = 2 \][/tex]
Thus, the vertex of this equation is [tex]\((-3, 2)\)[/tex].
2. Consider the equation [tex]\(y = 4x^2 - 24x + 38\)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = -24 \)[/tex]
- [tex]\( c = 38 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{-24}{2 \times 4} = \frac{24}{8} = 3 \][/tex]
The vertex here is not [tex]\((-3, 2)\)[/tex].
3. Consider the equation [tex]\(y = 4x^2 + 12x + 2\)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 12 \)[/tex]
- [tex]\( c = 2 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{12}{2 \times 4} = -\frac{12}{8} = -1.5 \][/tex]
The vertex here is not [tex]\((-3, 2)\)[/tex].
4. Consider the equation [tex]\(y = 4x^2 + 16x + 13\)[/tex]:
- [tex]\( a = 4 \)[/tex]
- [tex]\( b = 16 \)[/tex]
- [tex]\( c = 13 \)[/tex]
Calculate [tex]\( h \)[/tex]:
[tex]\[ h = -\frac{b}{2a} = -\frac{16}{2 \times 4} = -\frac{16}{8} = -2 \][/tex]
The vertex here is not [tex]\((-3, 2)\)[/tex].
After evaluating each equation, we confirm that the equation [tex]\(y = 4x^2 + 24x + 38\)[/tex] has a vertex at [tex]\((-3, 2)\)[/tex].
Thus, the equation that represents a graph with a vertex at [tex]\((-3, 2)\)[/tex] is:
[tex]\[ \boxed{1} \][/tex]
