The points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)
<h3>How to determine the points on the graph of the parabola other than the vertex and x-intercepts?</h3>
The equation of the parabola is given as:
t(x) = -x^2 + 4x - 3
The vertex of the parabola is the point where the graph is at the maximum or the minimum
While the x-intercept is the point where the graph crosses the x-axis i.e when y = 0
Having said that, we have the equation of the parabola to be
t(x) = -x^2 + 4x - 3
Set x = 5.
So, we have:
t(5) = -5^2 + 4 * 5 - 3
Evaluate the exponents
t(5) = -25 + 4 * 5 - 3
Evaluate the products
t(5) = -25 + 20 - 3
Evaluate the sum and the difference
t(5) = -8
Set x = 10.
So, we have:
t(10) = -10^2 + 4 * 10 - 3
Evaluate the exponents
t(10) = -100 + 4 * 10 - 3
Evaluate the products
t(10) = -100 + 40 - 3
Evaluate the sum and the difference
t(10) = -63
Hence, the points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)
Read more about parabola at:
brainly.com/question/4061870
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