<prompt> Fumana boholo ba palo e latelang ea quadratic equation.y 2x2 - 5x - 4  <response>: Instruction in English: Find the maximum of the following quadratic equation. . Response in English: A quadratic function in the form y ax2 bx c has its maximum or minimum at the vertex. The x-coordinate of the vertex can be found using the formula x -b2a. Substituting the values of a and b from the given quadratic equation, we have:x -b2ax --522x 54The y-coordinate of the vertex can be found by substituting the value of x back into the equation:y 2x2 - 5x - 4y 2542 - 554 - 4y -498Therefore, the vertex of the parabola, and hence the maximum point, is at 54, -498 . However, since a from above is positive a0, the parabola faces upward, meaning the vertex represents the minimum of the quadratic, and the maximum value of the quadratic function is infinity. . Response in Sesotho: Tshebetso ya quadratic ka sebopeho sa y ax2 bx c e na le boholo ba yona kapa bonyane ba yona hodimo. X-coordinate ea vertex e ka fumanoa ho sebelisoa foromo x -b2a. Ho fetola boleng ba a le b ho tswa ho quadratic equation e fanoeng, re na le: x -b2a x --522 x 54 Khokahano ea y ea vertex can fumanwa ka ho kenya boleng ba x hape ho equation: y 2x2 - 5x - 4 y 2542 - 554 - 4 y -498 Ka hona , vertex ea parabola, me kahoo ntlha e phahameng ka ho fetisisa, e ho 54, -498 . Leha ho le joalo, kaha a e tsoang holimo e nepahetse a0, parabola e shebane le holimo, ho bolelang hore vertex e emela bonyane ba quadratic, me boholo ba boleng ba mosebetsi oa quadratic ke bo sa feleng. 