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A quadratic function S^(9)G he function f(x)=-8x² vertex. Then plot another point on

User Core
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Alright, here's how we approach this problem:

1. We first need to understand the standard form of a quadratic equation, which is y = ax² + bx + c. The vertex of this form of equation is given by the point (-b/2a, f(-b/2a)), where a and b are coefficients of the equation.

2. Looking at our function, i.e., f(x) = -8x², we observe that there is no bx term in it, hence our 'b' value is 0. The coefficient 'a' is -8 here.

3. Now, we can utilize these values in the vertex point formula. We substitute the values a=-8 and b=0 into the vertex point formula, which gives us (0/(-2* -8), f(0/(-2*-8))), simplifying to (0, f(0)).

4. Substituting x=0 into our function f(x) = -8x², we get the second coordinate of the vertex point, which is -8*(0²)= 0. Thus, the vertex point from our calculation is (0, 0).

5. Now, to find another point that resides on this quadratic function, we can choose an arbitrary x value and substitute it into the equation. The result of this will be the corresponding y value for the x chosen.

6. Hence, we decide to pick x=1. Applying this value into the function f(x) = -8x² gives us the y coordinate -8*(1²), which is -8.

7. Conclusively, another point on the graph with the coordinates (1, -8).

So through our calculations, we managed to find the vertex of the function as the point (0, 0) and another point on the graph as (1, -8).

User Fei Xue
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