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I can help with Part A only. :( Is that okay or you need some more talented guy to solve all the three for you? :P
Okay, f(x) or y both are one and same thing, so you have : \(y = -16x^2 + 22x + 3\)
See, to find x-intercept, you put y = 0.. And to find y-intercept (if someone asks you to), then you put x = 0..
So, here you need to find x-intercept, so you put y = 0, so you have: \(y = -16x^2 + 22x + 3\) \(\implies 0 = -16x^2 + 22x + 3\)
All good till here?
sorta then what would you do
Then now you have got a quadratic equation to solve, solve for x there..
Part B: the vertex is: maximum with negative leading coefficient. minimum with positive leading coefficient. to find the vertex, complete the square.
Actually, x-intercept is the point where your graph cuts x-axis, and on x-axis, you have only x values, there y = 0..
told part B already.... ---------------------------------- For part C: Use the x-intercepts (you can also use the y-intercept which is simple to find -- optional) And for \("\)Justify that you can use the answers obtained in Part A and Part B to draw the graph.\("\) it is justified, but your graph of course won't look and be exact as on a graphing calculator.
Do you want to tell me right now, if your vertex is maximum or minimum ?