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Consider the following quadratic function. \[ g(x)=2 x^{2}+16 x+31 \] (a) Write the equation in the form \( g(x)=a(x-h)^{2}+k \). Then give the vertex of its graph. Writing in the form specified: \( g(x)=-2 x^{2}+16 x+31 \) Vertex: \( (4,63) \)

Ask by Ayala O'Quinn. in the United States
Mar 14,2025

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The quadratic function \( g(x) = 2x^2 + 16x + 31 \) can be rewritten in vertex form as \( g(x) = 2(x + 4)^2 - 1 \). The vertex of its graph is at \( (-4, -1) \).

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To rewrite the quadratic function \( g(x) = 2x^2 + 16x + 31 \) in vertex form, we can complete the square. First, factor out the coefficient of \( x^2 \) from the first two terms: \[ g(x) = 2(x^2 + 8x) + 31 \] Next, take half of the coefficient of \( x \) (which is 8), square it (getting 16), and add and subtract it inside the parentheses: \[ g(x) = 2(x^2 + 8x + 16 - 16) + 31 \] \[ g(x) = 2((x + 4)^2 - 16) + 31 \] \[ g(x) = 2(x + 4)^2 - 32 + 31 \] \[ g(x) = 2(x + 4)^2 - 1 \] Now the function is in vertex form \( g(x) = 2(x + 4)^2 - 1 \). The vertex of its graph is at the point \( (-4, -1) \). Oh, did you know that completing the square dates back to ancient civilizations? Babylonian mathematicians were already solving quadratic equations using geometric methods over 2000 years ago! They might not have had calculators, but their methods laid the foundation for algebra as we know it today. If you're planning to apply this knowledge practically, think about how quadratic functions appear in real life! Projectile motion is a perfect example; when you throw a ball, the path it follows is a parabola. Understanding the vertex helps to determine the peak height of the throw and the optimal angle to achieve the desired distance. It's math in action!

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