draw a diagram of the archway modeled by the equation y = -x² + 5x + 24. find and label the y - intercept…

draw a diagram of the archway modeled by the equation y = -x² + 5x + 24. find and label the y - intercept and the x - intercepts on the sketch. then find and label the width of the archway at its base and the height of the archway at its highest point, assuming the base of the archway is along the x - axis.

draw a diagram of the archway modeled by the equation y = -x² + 5x + 24. find and label the y - intercept and the x - intercepts on the sketch. then find and label the width of the archway at its base and the height of the archway at its highest point, assuming the base of the archway is along the x - axis.

Answer

Explanation:

Step1: Encontrar el intercepto en y

Sustituir (x = 0) en (y=-x^{2}+5x + 24). [y=-(0)^{2}+5(0)+24=24]

Step2: Encontrar los interceptos en x

Igualar (y = 0), es decir, resolver la ecuación (-x^{2}+5x + 24=0). Multiplicar por - 1 para obtener (x^{2}-5x - 24=0). Factorizar: ((x - 8)(x+3)=0). Entonces (x=8) o (x=-3).

Step3: Encontrar la anchura en la base

La anchura en la base es la distancia entre los dos valores de (x) donde (y = 0). (d=\vert8-(-3)\vert=11).

Step4: Encontrar el vértice

La (x) - coordenada del vértice de una parábola (y = ax^{2}+bx + c) es (x=-\frac{b}{2a}). Aquí (a=-1), (b = 5), entonces (x=-\frac{5}{2\times(-1)}=\frac{5}{2}). Sustituir (x=\frac{5}{2}) en la ecuación (y=-x^{2}+5x + 24): [y=-\left(\frac{5}{2}\right)^{2}+5\times\frac{5}{2}+24=-\frac{25}{4}+\frac{25}{2}+24=-\frac{25}{4}+\frac{50}{4}+24=\frac{-25 + 50}{4}+24=\frac{25}{4}+24=\frac{25+96}{4}=\frac{121}{4}=30.25]

Answer:

  • Intercepto en y: ((0,24))
  • Interceptos en x: ((-3,0)) y ((8,0))
  • Anchura en la base: (11)
  • Altura en el punto más alto: (30.25)