Given the function f(x)=3x^2−4x+5, determine the value of f(−2) and f(3). Then, find the x-intercepts of the function.
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Let's evaluate the quadratic function f of x equals 3 x squared minus 4 x plus 5. First, we need to find the values of f at x equals negative 2 and x equals 3.
To find f of negative 2, we substitute x equals negative 2 into the function. We get 3 times negative 2 squared, minus 4 times negative 2, plus 5. This equals 3 times 4, plus 8, plus 5, which gives us 12 plus 8 plus 5, or 25.
Similarly, for f of 3, we substitute x equals 3 into the function. We get 3 times 3 squared, minus 4 times 3, plus 5. This equals 3 times 9, minus 12, plus 5, which gives us 27 minus 12 plus 5, or 20.
Now, let's find the x-intercepts of the function. These are the points where the graph crosses the x-axis, or where f of x equals zero.
To find the x-intercepts, we set f of x equal to zero and solve the resulting quadratic equation: 3 x squared minus 4 x plus 5 equals zero. We'll use the quadratic formula, where a equals 3, b equals negative 4, and c equals 5.
First, we calculate the discriminant, delta, which equals b squared minus 4 a c. Substituting our values, we get negative 4 squared minus 4 times 3 times 5, which equals 16 minus 60, or negative 44.
Since the discriminant is negative, the quadratic equation has no real solutions. This means the function never crosses the x-axis, so there are no x-intercepts. Looking at the graph, we can see that the parabola stays above the x-axis for all values of x, with its lowest point at the vertex.
Let's summarize what we've learned. We evaluated the quadratic function f of x equals 3 x squared minus 4 x plus 5 at two specific points: f of negative 2 equals 25, and f of 3 equals 20. We found that this function has no x-intercepts because the discriminant is negative, which means the quadratic equation has no real solutions. The parabola opens upward and never crosses the x-axis. The vertex, or minimum point of the parabola, is approximately at the point (0.67, 6.33).