Answer:
Vertex (4.53 , 335.33) Zeroes are -0.048 and 9.11
Explanation:
For find the Vertex also know as the turning point/stationary point and the zeros also known as the x-intercepts of the equation we equate the equation to its complete square form which is
where (h,k) are the vertex point of the equation h corresponds to x coordinate and k corresponds to the y coordinate and a determines whether the graph is a maximum or a minimum. To find whether the graph is a maximum value of a should be < 0 which and if a's value is > 0 then the graph is a minimum. In short
maximum (n shaped)
minimum (u shaped
so now we equate, that is
![-16x^2+145x+7=a(x-h)^2+k](https://img.qammunity.org/2021/formulas/mathematics/college/uqn9edfv0z2hsj4k1j9git4ntm4525qj8e.png)
![-16x^2+145x+7=ax^2-2ahx+ah^2+k](https://img.qammunity.org/2021/formulas/mathematics/college/i9b1s5c1f1pl51dq5rksoyatn9fp5nb85t.png)
now we compare the coefficients of x^2
-
![-16=a](https://img.qammunity.org/2021/formulas/mathematics/college/qn1szx5i34enlhht1aehiyfckzpylxn85g.png)
now we compare the coefficients of x
![145=-2ah](https://img.qammunity.org/2021/formulas/mathematics/college/c5o6obx1eg99vfcjpfwv259qv777lbf91s.png)
![145=-2(-16)h](https://img.qammunity.org/2021/formulas/mathematics/college/7f7ydgxgk916jbzh4qng8tc31dft6a1wlv.png)
![145=32h](https://img.qammunity.org/2021/formulas/mathematics/college/e6ek8kllr632hq215gc8caqswzajb37vrw.png)
![4.53=h](https://img.qammunity.org/2021/formulas/mathematics/college/4nto697dzmbqxkxdfxcesituehqch7b0ly.png)
now we compare constants
![7=ah^2+k](https://img.qammunity.org/2021/formulas/mathematics/college/qmsxdhtlji7i7pc5b2eg67h717etqnyye8.png)
![7=(-16)(4.53)^2+k](https://img.qammunity.org/2021/formulas/mathematics/college/w9nsl32kjcn7yxmnke1d50qjz8elb3j6kl.png)
![7=-328.33+k](https://img.qammunity.org/2021/formulas/mathematics/college/fcvqorzvro5nxszvnlc1ftn30uswb8n8bv.png)
![335.33=k](https://img.qammunity.org/2021/formulas/mathematics/college/qv4is9kbrndt7bhos6ivzfxdk3czt71stb.png)
so now we the value of (h , k) that is (4.53 , 335.33)
that is our vertex and since the value of a is less than zero the graph is a maximum.
Now for the zeroes/x-intercepts
we need to find where the graph cuts the x-axis which means if the specific point where the curve cuts the x-axis that point's y coordinate should be zero or more like
.
and thus this point lies on the curve/equation as well because it satisfies it which means we can put this point into our given equation
when we put
in it y becomes 0 thus changing it to
![-16x^2+145x+7=0](https://img.qammunity.org/2021/formulas/mathematics/college/9udynoc5bggtaih94i9tnuyau849b23c62.png)
this becomes a simple quadratic equation where we can use the quadratic formula
±
![(√((145)^2-4(-16)(7)) )/(-32)](https://img.qammunity.org/2021/formulas/mathematics/college/fzw3e5vemi9zpi5nccmraddmvzcvdajjjz.png)
we get two values of x
and
![x=-(-145-146.54)/(-32)](https://img.qammunity.org/2021/formulas/mathematics/college/xcf5j1wigiieopdtdoexe3opa1z0ate0mz.png)
and
since there 2 values of x means the curve cuts the x-axis at two points we can even confirm our answers by using the desmos graphing calculator as well to check our vertex and zeroes.