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Use the function f(x) to answer the questions.

f(x) = −16x2 + 22x + 3

Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part B: Is the vertex of the graph of f(x) going to be a maximum or a minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

Sagot :

First graph

  • y=-16x^2+22x+3(Graph attached)

#A

Use quadratic formula

[tex]\\ \rm\rightarrowtail x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{484+192}}{-32}[/tex]

[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm\sqrt{676}}{-32}]/tex]

[tex]\\ \rm\rightarrowtail x=\dfrac{-22\pm 26}{-32}[/tex]

[tex]\\ \rm\rightarrowtail x=\dfrac{-1}{8}\:or\:\dfrac{3}{2}[/tex]

[tex]\\ \rm\rightarrowtail x=-0.125\:or\:1.5[/tex]

#B

Vertex will be maximum

Coordinates are

  • (0.688,10.563)

#C

  • We have to find x intercepts (Part A)
  • Then vertex(Part B)
  • Then we can graph it

View image Аноним
fieryanswererft

[tex]\sf f(x) = -16x^2 + 22x + 3[/tex]

Part A

To find x-intercepts, the f(x) will be 0

[tex]\sf \hookrightarrow -16x^2 + 22x + 3=0[/tex]

[tex]\sf \hookrightarrow -16x^2 + 24x -2x+ 3=0[/tex]

[tex]\sf \hookrightarrow -8x(2x-3)-1(2x- 3)=0[/tex]

[tex]\sf \hookrightarrow (-8x-1)(2x+ 3)=0[/tex]

[tex]\sf \hookrightarrow -8x-1=0 , \ 2x+ 3=0[/tex]

[tex]\sf \hookrightarrow x=-0.125 , \ x=1.5[/tex]

find y-intercept:

to find y-intercept, x will be 0

[tex]\sf f(x) = -16(0)^2 + 22(0) + 3[/tex]

[tex]\sf f(x) = 3[/tex]

Part B

To find vertex use the formula: x = -b/2(a)

[tex]\sf \rightarrow x = \dfrac{-22}{2(-16)}[/tex]

[tex]\sf \rightarrow x = 0.6875[/tex]

find y coordinates:

[tex]\sf f(x) = -16(0.6875)^2 + 22(0.6875) + 3[/tex]

[tex]\sf f(x) =10.5625[/tex]

coordinates of vertex: (0.6875, 10.5625)

  • [tex]\bold{\mathrm{If}\:a < 0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}}[/tex]
  • [tex]\bold{\mathrm{If}\:a > 0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}}[/tex]  

As its positive and greater than 0, the vertex will be Maximum.

part C

To draw the graph, simplify point the vertex on the graph. Then point both the x-intercepts. Also the y-intercept and draw the curve with free hand.

Graph attached:

View image fieryanswererft
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