## College Algebra (6th Edition)

$$f(x) = \frac{1}{2}(x + 3)^{2} -4$$
Standard form for any parabola is expressed as $f(x) = a(x-h)^{2} + k$, where the point value $(h,k)$ represents the vertex point. The parabola in question, then, can be expressed as: $$f(x) = a(x-(-3))^{2} + (-4) = a(x+3)^{2} - 4$$ where the value of $a$ is currently unknown. To calculate it, one must simply plug in the point value $(1,4)$ given by the exercise: $$f(1) = 4 = a((1)+3)^{2} - 4$$ $$4+4 = a(4)^{2}$$ $$8 = a(16)$$ $$\frac{8}{16} = \frac{1}{2} = a$$ The final expression in standard form comes to: $$f(x) = \frac{1}{2}(x + 3)^{2} -4$$