For this question, you will need to: 1- Choose the correct rewrite of \( y=5 x^{2}-2000 x+948 \) in vertex form. 2- Identify the correct vertex of the parabola. 3- Identify whether it is the maximum or minimum. Select one: \[ y=5\left(x^{2}-200\right)-199055 \text {; Vertex: }(200,-1990) \text {; Maximum } \] \( y=5(x-200)-199056 \); Vertex: \( (200,-199055) \); Maximum \( y=5(x-200)^{2}-199052 \); Vertex: \( (200,-199052) \); Minimum \( y=5(x-200)^{2}-1990 \); Vertex: \( (200,-199056) \); Minimum
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To convert the quadratic equation \( y=5 x^{2}-2000 x+948 \) into vertex form, we complete the square: Start with \( 5(x^{2}-400x) + 948 \). Taking \(-400\) and halving gives \(-200\). Squaring the result gives \(40000\). Thus, we adjust as follows: \[ y = 5((x-200)^{2} - 40000) + 948 \] \[ y = 5(x-200)^{2} - 199052 \] So the correct rewrite is: \[ y=5(x-200)^{2}-199052 \] The vertex of the parabola is \( (200, -199052) \), and since the coefficient of \( (x-200)^{2} \) is positive, this indicates the parabola opens upwards, making the vertex a minimum. The correct choice from your options is: \( y=5(x-200)^{2}-199052 \); Vertex: \( (200,-199052) \); Minimum. --- Looking back at quadratic functions, the vertex form is expressed as \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. This form is incredibly useful as it outlines the turning point of the parabola and simplifies determining the maximum or minimum value straight away. In real life, you can find applications for quadratic functions in various fields such as economics (profit maximization), sports (trajectory of objects), or architecture (designing arches). When working with parabolas, understanding the vertex helps in optimizing scenarios – like finding the best projecting angle for an object to reach its peak!
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